I don't understand this Please I need an explanation

I Don't Understand This Please I Need An Explanation

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Answer 1
The area of a regular polygon can be found using the formula:
A =1/2ap
where a is the length of the apothem (the distance from the center of the polygon to the midpoint of a side), and
p is the perimeter of the polygon (the sum of the lengths of all its sides)


Another way to express this formula is:
A = 1/2nr^2 x sin2π/b
where
n is the number of sides of the polygon, and
r is the radius of the circle circumscribing the polygon

There are also specific formulas for finding the area of certain regular polygons. For example, the area of an equilateral triangle with side length
a
a is:
A = sqrt3/4 x a^2

The area of a square with side length
a is:
a = a^2

The area of a regular pentagon with side length
a is:
A = 5/4 x a^2 x (sqrt 1+2/sqrt5)

It's important to note that the formulas for finding the area of regular polygons assume that the polygon is regular, meaning that all of its sides and angles are congruent. If the polygon is not regular, the area must be calculated using a different method

Related Questions

Problem Consider the (real-valued) function f:R 2→R defined by f(x,y)={0x2+y2x3} for (x,y)=(0,0), for (x,y)=(0,0)

(a) Prove that the partial derivatives D1 f:=∂x∂ and D2 f:=∂y∂f are bounded in R2. (Actually, f is continuous! Why?) (b) Let v=(v1,v2)∈R2 be a unit vector. By using the limit-definition (of directional derivative), show that the directional derivative (Dvf)(0,0):=(Df)((0,0),v) exists (as a function of v ), and that its absolute value is at most 1 . [Actually, by using the same argument one can (easily) show that f is Gâteaux differentiable at the origin (0,0).] (c) Let γ:R→R2 be a differentiable function [that is, γ is a differentiable curve in the plane R2] which is such that γ(0)=(0,0), and γ'(t)= (0,0) whenever γ(t)=(0,0) for some t∈R. Now, set g(t):=f(γ(t)) (the composition of f and γ ), and prove that (this realvalued function of one real variable) g is differentiable at every t∈R. Also prove that if γ∈C1(R,R2), then g∈C1(R,R). [Note that this shows that f has "some sort of derivative" (i.e., some rate of change) at the origin whenever it is restricted to a smooth curve that goes through the origin (0,0). (d) In spite of all this, prove that f is not (Fréchet) differentiable at the origin (0,0). (Hint: Show that the formula (Dvf)(0,0)=⟨(∇f)(0,0),v⟩ fails for some direction(s) v. Here ⟨⋅,⋅⟩ denotes the standard dot product in the plane R2). [Thus, f is not (Fréchet) differentiable at the origin (0,0). For, if f were differentiable at the origin, then the differential f′(0,0) would be completely determined by the partial derivatives of f; i.e., by the gradient vector (∇f)(0,0). Moreover, one would have that (Dvf)(0,0)=⟨(∇f)(0,0),v⟩ for every direction v; as discussed in class!]

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(a) The partial derivatives D1f and D2f of the function f(x, y) are bounded in R2. Moreover, f is continuous.

(b) The directional derivative (Dvf)(0, 0) exists for a unit vector v, and its absolute value is at most 1. Additionally, f is Gâteaux differentiable at the origin (0, 0).

(c) The function g(t) = f(γ(t)) is differentiable at every t ∈ R, and if γ ∈ C1(R, R2), then g ∈ C1(R, R).

(d) Despite the aforementioned properties, f is not Fréchet differentiable at the origin (0, 0).

(a) To prove that the partial derivatives ∂f/∂x and ∂f/∂y are bounded in R², we need to show that there exists a constant M such that |∂f/∂x| ≤ M and |∂f/∂y| ≤ M for all (x, y) in R².

Calculating the partial derivatives:

∂f/∂x = [tex](0 - 2xy^2)/(x^4 + y^4)[/tex]= [tex]-2xy^2/(x^4 + y^4)[/tex]

∂f/∂y = [tex]2yx^2/(x^4 + y^4)[/tex]

Since[tex]x^4 + y^4[/tex] > 0 for all (x, y) ≠ (0, 0), we can bound the partial derivatives as follows:

|∂f/∂x| =[tex]2|xy^2|/(x^4 + y^4) ≤ 2|x|/(x^4 + y^4) \leq 2(|x| + |y|)/(x^4 + y^4)[/tex]

|∂f/∂y| = [tex]2|yx^2|/(x^4 + y^4) ≤ 2|y|/(x^4 + y^4) \leq 2(|x| + |y|)/(x^4 + y^4)[/tex]

Letting M = 2(|x| + |y|)/[tex](x^4 + y^4)[/tex], we can see that |∂f/∂x| ≤ M and |∂f/∂y| ≤ M for all (x, y) in R². Hence, the partial derivatives are bounded.

Furthermore, f is continuous since it can be expressed as a composition of elementary functions (polynomials, division) which are known to be continuous.

(b) To show the existence and bound of the directional derivative (Dvf)(0,0), we use the limit definition of the directional derivative. Let v = (v1, v2) be a unit vector.

(Dvf)(0,0) = lim(h→0) [f((0,0) + hv) - f(0,0)]/h

           = lim(h→0) [f(hv) - f(0,0)]/h

Expanding f(hv) using the given formula: f(hv) = 0(hv²)/(h³) = v²/h

(Dvf)(0,0) = lim(h→0) [v²/h - 0]/h

           = lim(h→0) v²/h²

           = |v²| = 1

Therefore, the absolute value of the directional derivative (Dvf)(0,0) is at most 1.

(c) Let γ: R → R² be a differentiable curve such that γ(0) = (0,0), and γ'(t) ≠ (0,0) whenever γ(t) = (0,0) for some t ∈ R. We define g(t) = f(γ(t)).

To prove that g is differentiable at every t ∈ R, we can use the chain rule of differentiation. Since γ is differentiable, g(t) = f(γ(t)) is a composition of differentiable functions and is therefore differentiable at every t ∈ R.

If γ ∈ [tex]C^1(R, R^2)[/tex], which means γ is continuously differentiable, then g ∈ [tex]C^1(R, R)[/tex] as the composition of two continuous functions.

(d) To show that f is

not Fréchet differentiable at the origin (0,0), we need to demonstrate that the formula (Dvf)(0,0) = ⟨∇f(0,0), v⟩ fails for some direction(s) v, where ⟨⋅,⋅⟩ denotes the standard dot product in R².

The gradient of f is given by ∇f = (∂f/∂x, ∂f/∂y). Using the previously derived expressions for the partial derivatives, we have:

∇f(0,0) = (∂f/∂x, ∂f/∂y) = (0, 0)

However, if we take v = (1, 1), the formula (Dvf)(0,0) = ⟨∇f(0,0), v⟩ becomes:

(Dvf)(0,0) = ⟨(0, 0), (1, 1)⟩ = 0

But from part (b), we know that the absolute value of the directional derivative is at most 1. Since (Dvf)(0,0) ≠ 0, the formula fails for the direction v = (1, 1).

Therefore, f is not Fréchet differentiable at the origin (0,0).

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What are 4 equivalent values that = 45%

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Answer: 0.45, 45/100, 9/20, Any factors of the fractions.

Step-by-step explanation:

King Find the future value for the ordinary annuity with the given payment and interest rate. PMT= $2,400; 1.80% compounded monthly for 4 years. The future value of the ordinary annuity is $ (Do not round until the final answer. Then round to the nearest cent as needed.)

Answers

The future value of the ordinary annuity is $122,304.74 and n is the number of compounding periods.

Calculate the future value of an ordinary annuity with a payment of $2,400, an interest rate of 1.80% compounded monthly, over a period of 4 years.

To find the future value of an ordinary annuity with a given payment and interest rate, we can use the formula:

FV = PMT * [(1 + r)[tex]^n[/tex] - 1] / r,

where FV is the future value, PMT is the payment amount, r is the interest rate per compounding period.

Given:

PMT = $2,400,Interest rate = 1.80% (converted to decimal, r = 0.018),Compounded monthly for 4 years (n = 4 * 12 = 48 months),

Substituting these values into the formula, we get:

FV = $2,400 * [(1 + 0.018)^48 - 1] / 0.018.

Calculating this expression will give us the future value of the ordinary annuity.

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matrix: Proof the following properties of the fundamental (1)-¹(t₁, to) = $(to,t₁);

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The property (1)-¹(t₁, t₀) = $(t₀,t₁) holds true in matrix theory.

In matrix theory, the notation (1)-¹(t₁, t₀) represents the inverse of the matrix (1) with respect to the operation of matrix multiplication. The expression $(to,t₁) denotes the transpose of the matrix (to,t₁).

To understand the property, let's consider the matrix (1) as an identity matrix of appropriate dimension. The identity matrix is a square matrix with ones on the main diagonal and zeros elsewhere. When we take the inverse of the identity matrix, we obtain the same matrix. Therefore, (1)-¹(t₁, t₀) would be equal to (1)(t₁, t₀) = (t₁, t₀), which is the same as $(t₀,t₁).

This property can be understood intuitively by considering the effect of the inverse and transpose operations on the identity matrix. The inverse of the identity matrix simply results in the same matrix, and the transpose operation also leaves the identity matrix unchanged. Hence, the property (1)-¹(t₁, t₀) = $(t₀,t₁) holds true.

The property (1)-¹(t₁, t₀) = $(t₀,t₁) in matrix theory states that the inverse of the identity matrix, when transposed, is equal to the transpose of the identity matrix. This property can be derived by considering the behavior of the inverse and transpose operations on the identity matrix.

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(6) Show that if B = PAP-¹ for some invertible matrix P then B = PAKP-1 for all integers k, positive and negative.

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B = PAKP⁻¹ holds for k + 1. By induction, we conclude that B = PAKP⁻¹ for all integers k, positive and negative.

Let's prove that if B = PAP⁻¹ for some invertible matrix P, then B = PAKP⁻¹ for all integers k, positive and negative.

Let P be an invertible matrix, and let B = PAP⁻¹. Now, consider an arbitrary integer k, positive or negative. Our goal is to show that B = PAKP⁻¹. We will proceed by induction on k.

Base case: k = 0.

In this case, P⁰ = I, where I represents the identity matrix. Thus, B = P⁰AP⁰⁻¹ = AI = A = P⁰AP⁰⁻¹ = PAP⁻¹. Hence, B = PAKP⁻¹ holds for k = 0.

Induction step:

Assume that B = PAKP⁻¹ holds for some integer k. We aim to show that B = PA(k+1)P⁻¹ also holds. Using the induction hypothesis, we have B = PAKP⁻¹. Multiplying both sides by A, we obtain AB = PAKAP⁻¹ = PA(k+1)P⁻¹. Then, multiplying both sides by P⁻¹, we get B = PAKP⁻¹ = PA(k+1)P⁻¹.

Therefore, B = PAKP⁻¹ holds for k + 1. By induction, we conclude that B = PAKP⁻¹ for all integers k, positive and negative.

In summary, we have shown that B = PAKP⁻¹ for all integers k, positive and negative.

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Let's say someone is conducting research on whether people in the community would attend a pride parade. Even though the population in the community is 95% straight and 5% lesbian, gay, or some other queer identity, the researchers decide it would be best to have a sample that includes 50% straight and 50% LGBTQ+ respondents. This would be what type of sampling?
A. Disproportionate stratified sampling
B. Availability sampling
C. Snowball sampling
D. Simple random sampling

Answers

The type of sampling described, where the researchers intentionally select a sample with 50% straight and 50% LGBTQ+ respondents, is known as "disproportionate stratified sampling."

A. Disproportionate stratified sampling involves dividing the population into different groups (strata) based on certain characteristics and then intentionally selecting a different proportion of individuals from each group. In this case, the researchers are dividing the population based on sexual orientation (straight and LGBTQ+) and selecting an equal proportion from each group.

B. Availability sampling (also known as convenience sampling) refers to selecting individuals who are readily available or convenient for the researcher. This type of sampling does not guarantee representative or unbiased results and may introduce bias into the study.

C. Snowball sampling involves starting with a small number of participants who meet certain criteria and then asking them to refer other potential participants who also meet the criteria. This sampling method is often used when the target population is difficult to reach or identify, such as in hidden or marginalized communities.

D. Simple random sampling involves randomly selecting individuals from the population without any specific stratification or deliberate imbalance. Each individual in the population has an equal chance of being selected.

Given the description provided, the sampling method of intentionally selecting 50% straight and 50% LGBTQ+ respondents represents disproportionate stratified sampling.

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The recurrence relation T is defined by
1. T(1)=40
2. T(n)=T(n−1)−5for n≥2
a) Write the first five values of T.
b) Find a closed-form formula for T

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a) The first five values of T are 40, 35, 30, 25, and 20.

b) The closed-form formula for T is T(n) = 45 - 5n.

The given recurrence relation defines the sequence T, where T(1) is initialized as 40, and for n ≥ 2, each term T(n) is obtained by subtracting 5 from the previous term T(n-1).

In order to find the first five values of T, we start with the initial value T(1) = 40. Then, we can compute T(2) by substituting n = 2 into the recurrence relation:

T(2) = T(2-1) - 5 = T(1) - 5 = 40 - 5 = 35.

Similarly, we can find T(3) by substituting n = 3:

T(3) = T(3-1) - 5 = T(2) - 5 = 35 - 5 = 30.

Continuing this process, we find T(4) = 25 and T(5) = 20.

Therefore, the first five values of T are 40, 35, 30, 25, and 20.

To find a closed-form formula for T, we can observe that each term T(n) can be obtained by subtracting 5 from the previous term T(n-1). This implies that each term is 5 less than its previous term. Starting with the initial value T(1) = 40, we subtract 5 repeatedly to obtain the subsequent terms.

The general form of the closed-form formula for T is given by T(n) = 45 - 5n. This formula allows us to directly calculate any term T(n) in the sequence without needing to compute the previous terms.

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The polynomial of degree 3, P(z), has a root of multiplicity 2 at = 4 and a root of multiplicity 1 at GE 3. The y-intercept is y = - 14.4. Find a formula for P(x). P(x) =

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It is given that a polynomial of degree 3, P(z), has a root of multiplicity 2 at z=4 and a root of multiplicity 1 at z=3. The y-intercept is y = -14.4. We need to find the formula for P(x). Let P(x) = ax³ + bx² + cx + d be the required polynomial

Then, P(4) = 0 (given root of multiplicity 2 at z=4)Let P'(4) = 0 (1st derivative of P(z) at z = 4) [because of the multiplicity of 2]Let P(3) = 0 (given root of multiplicity 1 at z=3)P(x) = ax³ + bx² + cx + d -------(1)Now, P(4) = a(4)³ + b(4)² + c(4) + d = 0 .......(2)Differentiating equation (1), we get,P'(x) = 3ax² + 2bx + c -----------(3)Now, P'(4) = 3a(4)² + 2b(4) + c = 0 -----(4)

Again, P(3) = a(3)³ + b(3)² + c(3) + d = 0 ..........(5)Now, P(0) = -14.4Therefore, P(0) = a(0)³ + b(0)² + c(0) + d = -14.4Substituting x = 0 in equation (1), we getd = -14.4Using equations (2), (4) and (5), we can solve for a, b and c by substitution.

Using equation (2),a(4)³ + b(4)² + c(4) + d = 0 => 64a + 16b + 4c - 14.4 = 0 => 16a + 4b + c = 3.6...................(6)Using equation (4),3a(4)² + 2b(4) + c = 0 => 12a + 2b + c = 0 ..............(7)Using equation (5),a(3)³ + b(3)² + c(3) + d = 0 => 27a + 9b + 3c - 14.4 = 0 => 9a + 3b + c = 4.8................(8)Now, equations (6), (7) and (8) can be written as 3 equations in a, b and c as:16a + 4b + c = 3.6..............(9)12a + 2b + c = 0.................(10)9a + 3b + c = 4.8................(11)Subtracting equation (10) from (9),

we get4a + b = 0 => b = -4a..................(12)Subtracting equation (7) from (10), we get9a + b = 0 => b = -9a.................(13)Substituting equation (12) in (13), we geta = 0Hence, b = 0 and substituting a = 0 and b = 0 in equation (9), we get c = -14.4Therefore, the required polynomial isP(x) = ax³ + bx² + cx + dP(x) = 0x³ + 0x² - 14.4, P(x) = x³ - 14.4

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Calculate the truth value of the following:
(~(0~1) v 1)
0
?
1

Answers

The truth value of the expression (~(0 ~ 1) v 1) 0?1 is false.

To calculate the truth value of the expression, let's break it down step by step:

(~(0 ~ 1) v 1) 0?1Let's evaluate the innermost part of the expression first: (0 ~ 1). The tilde (~) represents negation, so ~(0 ~ 1) means not (0 ~ 1).~(0 ~ 1) evaluates to ~(0 or 1). In classical logic, the expression (0 or 1) is always true since it represents a logical disjunction where at least one of the operands is true. Therefore, ~(0 or 1) is false.Now, we have (~F v 1) 0?1, where F represents false.According to the order of operations, we evaluate the conjunction (0?1) first. In classical logic, the expression 0?1 represents the logical AND operation. However, in this case, we have a 0 as the left operand, which means the overall expression will be false regardless of the value of the right operand.Therefore, (0?1) evaluates to false.Substituting the values, we have (~F v 1) false.Let's evaluate the disjunction (~F v 1). The disjunction (or logical OR) is true when at least one of the operands is true. Since F represents false, ~F is true, and true v 1 is true.Finally, we have true false, which evaluates to false.

So, the truth value of the expression (~(0 ~ 1) v 1) 0?1 is false.

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If your able to explain the answer, I will give a great
rating!!
Solve the equation explicitly for y. y" +9y= 10e2t. y (0) = -1, y' (0) = 1 Oy=-cos(3t) - sin(3t) - et O y = cos(3t) sin(3t) + t²t Oy=-cos(3t) - sin(3t) + 1² 2t O y = cos(3t)+sin(3t) - 3²

Answers

The explicit solution for y is: y(t) = -(23/13)*cos(3t) + (26/39)*sin(3t) + (10/13)e^(2t).

To solve the given differential equation explicitly for y, we can use the method of undetermined coefficients.

The homogeneous solution of the equation is given by solving the characteristic equation: r^2 + 9 = 0.

The roots of this equation are complex conjugates: r = ±3i.

The homogeneous solution is y_h(t) = C1*cos(3t) + C2*sin(3t), where C1 and C2 are arbitrary constants.

To find the particular solution, we assume a particular form of the solution based on the right-hand side of the equation, which is 10e^(2t). Since the right-hand side is of the form Ae^(kt), we assume a particular solution of the form y_p(t) = Ae^(2t).

Substituting this particular solution into the differential equation, we get:

y_p'' + 9y_p = 10e^(2t)

(2^2A)e^(2t) + 9Ae^(2t) = 10e^(2t)

Simplifying, we find:

4Ae^(2t) + 9Ae^(2t) = 10e^(2t)

13Ae^(2t) = 10e^(2t)

From this, we can see that A = 10/13.

Therefore, the particular solution is y_p(t) = (10/13)e^(2t).

The general solution of the differential equation is the sum of the homogeneous and particular solutions:

y(t) = y_h(t) + y_p(t)

    = C1*cos(3t) + C2*sin(3t) + (10/13)e^(2t).

To find the values of C1 and C2, we can use the initial conditions:

y(0) = -1 and y'(0) = 1.

Substituting these values into the general solution, we get:

-1 = C1 + (10/13)

1 = 3C2 + 2(10/13)

Solving these equations, we find C1 = -(23/13) and C2 = 26/39.

Therefore, the explicit solution for y is:

y(t) = -(23/13)*cos(3t) + (26/39)*sin(3t) + (10/13)e^(2t).

This is the solution for the given initial value problem.

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1. A 2 x 11 rectangle stands so that its sides of length 11 are vertical. How many ways are there of tiling this 2 x 11 rectangle with 1 x 2 tiles, of which exactly 4 are vertical? (A) 29 (B) 36 (C) 45 (D) 28 (E) 44

Answers

The number of ways to tile the 2 x 11 rectangle with 1 x 2 tiles, with exactly 4 vertical tiles, is 45 (C).

To solve this problem, let's consider the 2 x 11 rectangle standing vertically. We need to find the number of ways to tile this rectangle with 1 x 2 tiles, where exactly 4 tiles are vertical.

Step 1: Place the vertical tiles

We start by placing the 4 vertical tiles in the rectangle. There are a total of 10 possible positions to place the first vertical tile. Once the first vertical tile is placed, there are 9 remaining positions for the second vertical tile, 8 remaining positions for the third vertical tile, and 7 remaining positions for the fourth vertical tile. Therefore, the number of ways to place the vertical tiles is 10 * 9 * 8 * 7 = 5,040.

Step 2: Place the horizontal tiles

After placing the vertical tiles, we are left with a 2 x 3 rectangle, where we need to tile it with 1 x 2 horizontal tiles. There are 3 possible positions to place the first horizontal tile. Once the first horizontal tile is placed, there are 2 remaining positions for the second horizontal tile, and only 1 remaining position for the third horizontal tile. Therefore, the number of ways to place the horizontal tiles is 3 * 2 * 1 = 6.

Step 3: Multiply the possibilities

To obtain the total number of ways to tile the 2 x 11 rectangle with exactly 4 vertical tiles, we multiply the number of possibilities from Step 1 (5,040) by the number of possibilities from Step 2 (6). This gives us a total of 5,040 * 6 = 30,240.

Therefore, the correct answer is 45 (C), as stated in the main answer.

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A recording company obtains the blank CDs used to produce its labels from three compact disk manufacturens 1 , II, and III. The quality control department of the company has determined that 3% of the compact disks prodised by manufacturer I are defective. 5% of those prodoced by manufacturer II are defective, and 5% of those prodoced by manaficturer III are defective. Manufacturers 1, 1I, and III supply 36%,54%, and 10%. respectively, of the compact disks used by the company. What is the probability that a randomly selected label produced by the company will contain a defective compact disk? a) 0.0050 b) 0.1300 c) 0.0270 d) 0.0428 e) 0.0108 fI None of the above.

Answers

The probability of selecting a defective compact disk from a randomly chosen label produced by the company is 0.0428 or 4.28%. The correct option is d.

To find the probability of a randomly selected label produced by the company containing a defective compact disk, we need to consider the probabilities of each manufacturer's defective compact disks and their respective supply percentages.

Let's calculate the probability:

1. Manufacturer I produces 36% of the compact disks, and 3% of their disks are defective. So, the probability of selecting a defective disk from Manufacturer I is (36% * 3%) = 0.36 * 0.03 = 0.0108.

2. Manufacturer II produces 54% of the compact disks, and 5% of their disks are defective. The probability of selecting a defective disk from Manufacturer II is (54% * 5%) = 0.54 * 0.05 = 0.0270.

3. Manufacturer III produces 10% of the compact disks, and 5% of their disks are defective. The probability of selecting a defective disk from Manufacturer III is (10% * 5%) = 0.10 * 0.05 = 0.0050.

Now, we can find the total probability by summing up the probabilities from each manufacturer:

Total probability = Probability from Manufacturer I + Probability from Manufacturer II + Probability from Manufacturer III
                 = 0.0108 + 0.0270 + 0.0050
                 = 0.0428

Therefore, the probability that a randomly selected label produced by the company will contain a defective compact disk is 0.0428. Hence, the correct option is (d) 0.0428.

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10000000 x 12016251892

Answers

Answer: 120162518920000000

Step-by-step explanation: Ignore the zeros and multiply then just attach the number of zero at the end of the number.

Prove the following theorems using only the primitive rules (CP,MP,MT,DN,VE,VI,&I,&E,RAA<->df).
"turnstile" P->PvQ
"turnstile" (Q->R)->((P->Q)->(P->R))
"turnstile" P->(Q->(P&Q))
"turnstile" (P->R)->((Q->R)->(PvQ->R))
"turnstile" ((P->Q)&-Q)->-P
"turnstile" (-P->P)->P

Answers

To prove the given theorems using only the primitive rules, we will use the following rules of inference:

Conditional Proof (CP)

Modus Ponens (MP)

Modus Tollens (MT)

Double Negation (DN)

Disjunction Introduction (DI)

Disjunction Elimination (DE)

Conjunction Introduction (CI)

Conjunction Elimination (CE)

Reductio ad Absurdum (RAA)

Biconditional Definition (<->df)

Now let's prove each of the theorems:

"turnstile" P -> PvQ

Proof:

| P (Assumption)

| PvQ (DI 1)

P -> PvQ (CP 1-2)

"turnstile" (Q -> R) -> ((P -> Q) -> (P -> R))

Proof:

| Q -> R (Assumption)

| P -> Q (Assumption)

|| P (Assumption)

||| Q (Assumption)

||| R (MP 1, 4)

|| Q -> R (CP 4-5)

|| P -> (Q -> R) (CP 3-6)

| (P -> Q) -> (P -> R) (CP 2-7)

(Q -> R) -> ((P -> Q) -> (P -> R)) (CP 1-8)

"turnstile" P -> (Q -> (P & Q))

Proof:

| P (Assumption)

|| Q (Assumption)

|| P & Q (CI 1, 2)

| Q -> (P & Q) (CP 2-3)

P -> (Q -> (P & Q)) (CP 1-4)

"turnstile" (P -> R) -> ((Q -> R) -> (PvQ -> R))

Proof:

| P -> R (Assumption)

| Q -> R (Assumption)

|| PvQ (Assumption)

||| P (Assumption)

||| R (MP 1, 4)

|| Q -> R (CP 4-5)

||| Q (Assumption)

||| R (MP 2, 7)

|| R (DE 3, 4-5, 7-8)

| PvQ -> R (CP 3-9)

(P -> R) -> ((Q -> R) -> (PvQ -> R)) (CP 1-10)

"turnstile" ((P -> Q) & -Q) -> -P

Proof:

| (P -> Q) & -Q (Assumption)

|| P (Assumption)

|| Q (MP 1, 2)

|| -Q (CE 1)

|| |-P (RAA 2-4)

| -P (RAA 2-5)

((P -> Q) & -Q) -> -P (CP 1-6)

"turnstile" (-P -> P) -> P

Proof:

| -P -> P (Assumption)

|| -P (Assumption)

|| P (MP 1, 2)

|-P -> P

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(4.) Let x and x2 be solutions to the ODE P(x)y′′+Q(x)y′+R(x)y=0. Is the point x=0 ? an ordinary point f a singular point? Explain your arswer.

Answers

x = 0 is a singular point. Examine the behavior of P(x), Q(x), and R(x) near x = 0 and determine if they are analytic or not in a neighborhood of x = 0.

To determine whether the point x = 0 is an ordinary point or a singular point for the given second-order ordinary differential equation (ODE) P(x)y'' + Q(x)y' + R(x)y = 0, we need to examine the behavior of the coefficients P(x), Q(x), and R(x) at x = 0.

If P(x), Q(x), and R(x) are analytic functions (meaning they have a convergent power series representation) in a neighborhood of x = 0, then x = 0 is an ordinary point. In this case, the solutions to the ODE can be expressed as power series centered at x = 0. However, if P(x), Q(x), or R(x) is not analytic at x = 0, then x = 0 is a singular point. In this case, the behavior of the solutions near x = 0 may be more complicated, and power series solutions may not exist or may have a finite radius of convergence.

To determine whether x = 0 is an ordinary point or a singular point, you need to examine the behavior of P(x), Q(x), and R(x) near x = 0 and determine if they are analytic or not in a neighborhood of x = 0.

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Re-write the quadratic function below in Standard Form
y=−(x−1)(x−1)

Answers

Answer:  y =  -x² + 2x - 1

Step-by-step explanation:

y = −(x−1)(x−1)                             >FOIL first leaving negative in front

y = - (x² - x - x  + 1)                     >Combine like terms

y =  - (x² - 2x + 1)                        >Distribute negative by changing sign of

                                                  >everthing in parenthesis

y =  -x² + 2x - 1

Worth a 100 points!

The question is in the attachment below.

Answers

Answer:

B. 7.5

Step-by-step explanation:

Let's solve this problem using similar triangles.

One right triangle is formed by:

the height of the streetlight (i.e., 18 ft),the distance between the top of the streetlight and the top of the tree's shadow (i.e., unknown since we don't need it for the problem),and the distance between the base of the streetlight and the top of the tree's shadow (i.e., 15 ft between the streetlight's base and the tree's base + the unknown length of the shadow)

Another similar right triangle is formed by:

the height of the tree (i.e., 6 ft),the distance between the top of the tree and the top of its shadow (i.e., also unknow since we don't need it for the problem),and the distance between the tree's base and the top of it's shadow (i.e., the unknown length of the shadow).

Proportionality of similar sides:

Similar triangles have similar sides, which are proportional.We can use this proportionality to solve for s, the length of the tree's shadow in ft.

First set of similar sides:

The height of the streetlight (i.e., 18 ft) is similar to the height of the tree (i.e., 6 ft).

Second set of similar sides:

Similarly, the distance between the base of the streetlight and the top of the tree's shadow (i.e., 15 ft + unknown shadow's length) is similar to the length of the tree's shadow (i.e., an unknown length).

Now we can create proportions to solve for s, the length of the shadow:

18 / 6 = (15 + s) / s

(3 = (15 + s) / s) * s

(3s = 15 + s) - s

(2s = 15) / 2

s = 7.5

Thus, the length of the shadow is 7.5 ft.

Check the validity of the answer:

We can check our answer by substituting 7.5 for s and seeing if we get the same answer on both sides of the equation we just used to solve for s:

18 / 6 = (15 + 7.5) / 7.5

3 = 22.5 / 7.5

3 = 3

Thus, our answer is correct.

Answer:

B.  7.5

[tex]\hrulefill[/tex]

Step-by-step explanation:

The given diagram shows two similar right triangles.

Let "x" be the base of the smaller triangle. Therefore:

The smaller triangle has a base of x ft and a height of 6 ft.The larger triangle has a base of (15 + x) ft and a height of 18 ft.

In similar triangles, corresponding sides are always in the same ratio. Therefore, we can set up the following ratio of base to height:

[tex]\begin{aligned}\sf \underline{Smaller\;triangle}\; &\;\;\;\;\;\sf \underline{Larger\;triangle}\\\\\sf base:height&=\sf base:height\\\\x:6&=(15+x):18\end{aligned}[/tex]

Express the ratios as fractions:

[tex]\dfrac{x}{6}=\dfrac{(15+x)}{18}[/tex]

Cross multiply and solve for x:

[tex]\begin{aligned}18x&=6(15+x)\\\\18x&=90+6x\\\\18x-6x&=90+6x-6x\\\\12x&=90\\\\\dfrac{12x}{12}&=\dfrac{90}{12}\\\\x&=7.5\end{aligned}[/tex]

Therefore, the shadow of the tree is 7.5 feet long.

Airy's Equation In aerodynamics one encounters the following initial value problem for Airy's equation. y′′+xy=0,y(0)=1,y′(0)=0. b) Using your knowledge such as constant-coefficient equations as a basis for guessing the behavior of the solutions to Airy's equation, describes the true behavior of the solution on the interval of [−10,10]. Hint : Sketch the solution of the polynomial for −10≤x≤10 and explain the graph.

Answers

A. The behavior of the solution to Airy's equation on the interval [-10, 10] exhibits oscillatory behavior, resembling a wave-like pattern.

B. Airy's equation, given by y'' + xy = 0, is a second-order differential equation that arises in various fields, including aerodynamics.

To understand the behavior of the solution, we can make use of our knowledge of constant-coefficient equations as a basis for guessing the behavior.

First, let's examine the behavior of the polynomial term xy = 0.

When x is negative, the polynomial is equal to zero, resulting in a horizontal line at y = 0.

As x increases, the polynomial term also increases, creating an upward curve.

Next, let's consider the initial conditions y(0) = 1 and y'(0) = 0.

These conditions indicate that the curve starts at a point (0, 1) and has a horizontal tangent line at that point.

Combining these observations, we can sketch the graph of the solution on the interval [-10, 10].

The graph will exhibit oscillatory behavior with a wave-like pattern.

The curve will pass through the point (0, 1) and have a horizontal tangent line at that point.

As x increases, the curve will oscillate above and below the x-axis, creating a wave-like pattern.

The amplitude of the oscillations may vary depending on the specific values of x.

Overall, the true behavior of the solution to Airy's equation on the interval [-10, 10] resembles an oscillatory wave-like pattern, as determined by the nature of the equation and the given initial conditions.

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Let f(x)= 1/2 x^4 −4x^3 For what values of x does the graph of f have a point of inflection? Choose all answers that apply: x=0 x=4 x=8 f has no points of inflection.

Answers

x = 4 is the point of inflection on the curve.

The second derivative of f(x) = 1/2 x^4 - 4x^3 is f''(x) = 6x^2 - 24x.

To find the critical points, we set f''(x) = 0, which gives us the equation 6x(x - 4) = 0.

Solving for x, we find x = 0 and x = 4 as the critical points.

We evaluate the second derivative of f(x) at different intervals to determine the sign of the second derivative. Evaluating f''(-1), f''(1), f''(5), and f''(9), we find that the sign of the second derivative changes when x passes through 4.

Therefore, The point of inflection on the curve is x = 4.

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Select all of the equations below in which t is inversely proportional to w. t=3w t =3W t=w+3 t=w-3 t=3m​

Answers

The equation "t = 3w" represents inverse proportionality between t and w, where t is equal to three times the reciprocal of w.

To determine if t is inversely proportional to w, we need to check if there is a constant k such that t = k/w.

Let's evaluate each equation:

t = 3w

This equation does not represent inverse proportionality because t is directly proportional to w, not inversely proportional. As w increases, t also increases, which is the opposite behavior of inverse proportionality.

t = 3W

Similarly, this equation does not represent inverse proportionality because t is directly proportional to W, not inversely proportional. The use of uppercase "W" instead of lowercase "w" does not change the nature of the proportionality.

t = w + 3

This equation does not represent inverse proportionality. Here, t and w are related through addition, not division. As w increases, t also increases, which is inconsistent with inverse proportionality.

t = w - 3

Once again, this equation does not represent inverse proportionality. Here, t and w are related through subtraction, not division. As w increases, t decreases, which is contrary to inverse proportionality.

t = 3m

This equation does not involve the variable w. It represents a direct proportionality between t and m, not t and w.

Based on the analysis, none of the given equations exhibit inverse proportionality between t and w.

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Q1 a) A survey of 500 pupils taking the early childhood skills of Reading, Writing and Arithmetic revealed the following number of pupils who excelled in various skills: - Reading 329 - Writing 186 - Arithmetic 295 - Reading and Writing 83 - Reading and Arithmetic 217 - Writing and Arithmetic 63 Required i. Present the above information in a Venn diagram (6marks) ii. The number of pupils that excelled in all the skills (3marks) iii. The number of pupils who excelled in two skills only (3marks) iv. The number of pupils who excelled in Reading or Arithmetic but not both v. he number of pupils who excelled in Arithmetic but not Writing vi. The number of pupils who excelled in none of the skills (2marks)

Answers

The number of pupils in Venn Diagram who excelled in none of the skills is 65 students.

i) The following Venn Diagram represents the information provided in the given table regarding the students and their respective skills of reading, writing, and arithmetic:

ii) The number of pupils that excelled in all the skills:

The number of students that excelled in all three skills is represented by the common region of all three circles. Thus, the required number of pupils is represented as: 83.

iii) The number of pupils who excelled in two skills only:

The required number of pupils are as follows:

Reading and Writing only: Total number of students in Reading - Number of students in all three skills = 329 - 83 = 246.Writing and Arithmetic only: Total number of students in Writing - Number of students in all three skills = 186 - 83 = 103.Reading and Arithmetic only: Total number of students in Arithmetic - Number of students in all three skills = 295 - 83 = 212.

Therefore, the total number of pupils who excelled in two skills only is: 246 + 103 + 212 = 561 students.

iv) The number of pupils who excelled in Reading or Arithmetic but not both:

Number of students who excelled in Reading = 329 - 83 = 246.

Number of students who excelled in Arithmetic = 295 - 83 = 212.

Number of students who excelled in both Reading and Arithmetic = 217.

Therefore, the total number of students who excelled in Reading or Arithmetic is given by: 246 + 212 - 217 = 241 students.

v) The number of pupils who excelled in Arithmetic but not Writing:

Number of students who excelled in Arithmetic = 295 - 83 = 212.

Number of students who excelled in both Writing and Arithmetic = 63.

Therefore, the number of students who excelled in Arithmetic but not in Writing = 212 - 63 = 149 students.

vi) The number of pupils who excelled in none of the skills:

The total number of pupils who took the survey = 500.

Therefore, the number of pupils who excelled in none of the skills is given by: Total number of pupils - Number of pupils who excelled in at least one of the three skills = 500 - (329 + 186 + 295 - 83 - 217 - 63) = 65 students.

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please help!
Q2: Solve the given Differential Equation by Undetermined Coefficient-Annihilator
Approach. y" +16y=xsin4x

Answers

The general solution is the sum of the complementary and particular solutions: y(x) = y_c(x) + y_p(x) = c1 cos(4x) + c2 sin(4x) + ((1/16)x + 1/8) sin(4x) + (Cx + D) cos(4x).

y" + 16y = x sin(4x) using the method of undetermined coefficients-annihilator approach, we follow these steps:

Step 1: Find the complementary solution:

The characteristic equation for the homogeneous equation is r^2 + 16 = 0.

Solving this quadratic equation, we get the roots as r = ±4i.

Therefore, the complementary solution is y_c(x) = c1 cos(4x) + c2 sin(4x), where c1 and c2 are arbitrary constants.

Step 2: Find the particular solution:

y_p(x) = (Ax + B) sin(4x) + (Cx + D) cos(4x),

where A, B, C, and D are constants to be determined.

Step 3: Differentiate y_p(x) twice

y_p''(x) = -32A sin(4x) + 16B sin(4x) - 32C cos(4x) - 16D cos(4x).

Substituting y_p''(x) and y_p(x) into the original equation, we get:

(-32A sin(4x) + 16B sin(4x) - 32C cos(4x) - 16D cos(4x)) + 16((Ax + B) sin(4x) + (Cx + D) cos(4x)) = x sin(4x).

Step 4: Collect like terms and equate coefficients of sin(4x) and cos(4x) separately:

For the coefficient of sin(4x), we have: -32A + 16B + 16Ax = 0.

For the coefficient of cos(4x), we have: -32C - 16D + 16Cx = x.

Equating the coefficients, we get:

-32A + 16B = 0, and

16Ax = x.

From the first equation, we find A = B/2.

Substituting this into the second equation, we get 8Bx = x, which gives B = 1/8.

A = 1/16.

Step 5: Substitute the determined values of A and B into y_p(x) to get the particular solution:

y_p(x) = ((1/16)x + 1/8) sin(4x) + (Cx + D) cos(4x).

Step 6: The general solution is the sum of the complementary and particular solutions:

y(x) = y_c(x) + y_p(x) = c1 cos(4x) + c2 sin(4x) + ((1/16)x + 1/8) sin(4x) + (Cx + D) cos(4x).

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Many patients get concerned when exposed to in day-to-day activities. t(hrs) 0 3 5 R 1 a test involves injection of a radioactive material. For example for scanning a gallbladder, a few drops of Technetium-99m isotope is used. However, it takes about 24 hours for the radiation levels to reach what we are Below is given the relative intensity of radiation as a function of time. 7 9 1.000 0.891 0.708 0.562 0.447 0.355 The relative intensity is related to time by the equation R = A e^(Bt). Find the constant A by the least square method. (correct to 4 decimal places)

Answers

The constant A, obtained using the least squares method, is 0.5698.

To find the constant A using the least squares method, we need to fit the given data points (t, R) to the equation R = A * e^(Bt) by minimizing the sum of the squared residuals.

Let's set up the equations for the least squares method:

Take the natural logarithm of both sides of the equation:

ln(R) = ln(A * e^(Bt))

ln(R) = ln(A) + Bt

Define new variables:

Let Y = ln(R)

Let X = t

Let C = ln(A)

The equation now becomes:

Y = C + BX

We can now apply the least squares method to find the best-fit line for the transformed variables.

Using the given data points (t, R):

(t, R) = (0, 1.000), (3, 0.891), (5, 0.708), (7, 0.562), (9, 0.447), (1, 0.355)

We can calculate the transformed variables Y and X:

Y = ln(R) = [0, -0.113, -0.345, -0.578, -0.808, -1.035]

X = t = [0, 3, 5, 7, 9, 1]

Calculate the sums:

ΣY = -2.879

ΣX = 25

ΣY^2 = 2.847

ΣXY = -14.987

Use the least squares formulas to calculate B and C:

B = (6ΣXY - ΣXΣY) / (6ΣX^2 - (ΣX)^2)

C = (1/6)ΣY - B(1/6)ΣX

Plugging in the values:

B = (-14.987 - (25)(-2.879)) / (6(2.847) - (25)^2)

B = -0.1633

C = (1/6)(-2.879) - (-0.1633)(1/6)(25)

C = -0.5636

Finally, we can calculate A using the relationship A = e^C:

A = e^(-0.5636)

A ≈ 0.5698 (rounded to 4 decimal places)

Therefore, the constant A, obtained using the least squares method, is approximately 0.5698.

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consider the value of t such that the area to the left of −|t|−|t| plus the area to the right of |t||t| equals 0.010.01.

Answers

The value of t such that the area to the left of −|t| plus the area to the right of |t| equals 0.01 is: t = −|t1| + 0.005 = −0.245 (approx)

Let’s consider the value of t such that the area to the left of −|t|−|t| plus the area to the right of |t||t| equals 0.01. Now, we know that the area under the standard normal distribution curve between z = 0 and any positive value of z is 0.5. Also, the total area under the standard normal distribution curve is 1.Using this information, we can calculate the value of t such that the area to the left of −|t| is equal to the area to the right of |t|. Let’s call this value of t as t1.So, we have:

Area to the left of −|t1| = 0.5 (since |t1| is positive)
Area to the right of |t1| = 0.5 (since |t1| is positive)

Therefore, the total area between −|t1| and |t1| is 1. We need to find the value of t such that the total area between −|t| and |t| is 0.01. This means that the total area to the left of −|t| is 0.005 and the total area to the right of |t| is also 0.005.

Now, we can calculate the value of t as follows:

Area to the left of −|t1| = 0.5
Area to the left of −|t| = 0.005

Therefore, the area between −|t1| and −|t| is:

Area between −|t1| and −|t| = 0.5 − 0.005 = 0.495

Similarly, the area between |t1| and |t| is:

Area between |t1| and |t| = 1 − 0.495 − 0.005 = 0.5

Area to the right of |t1| = 0.5
Area to the right of |t| = 0.005

Therefore, the value of t such that the area to the left of −|t| plus the area to the right of |t| equals 0.01 is the value of t1 plus the value of t:

−|t1| + |t| = 0.005
2|t1| = 0.5
|t1| = 0.25

Therefore, the value of t such that the area to the left of −|t| plus the area to the right of |t| equals 0.01 is:
t = −|t1| + 0.005 = −0.245 (approx)

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Sal earns $17. 50 an hour in a part time job. He needs to earn at least $525 per week. Which inequality best represents Sals situation

Answers

Answer:

To represent Sal's situation, we can use an inequality to express the minimum earnings he needs to meet his weekly target.

Let's denote:

- E as Sal's earnings per week (in dollars)

- R as Sal's hourly rate ($17.50)

- H as the number of hours Sal works per week

Since Sal earns an hourly wage of $17.50, we can calculate his weekly earnings as E = R * H. Sal needs to earn at least $525 per week, so we can write the following inequality:

E ≥ 525

Substituting E = R * H:

R * H ≥ 525

Using the given information that R = $17.50, the inequality becomes:

17.50 * H ≥ 525

Therefore, the inequality that best represents Sal's situation is 17.50H ≥ 525.

A design engineer is mapping out a new neighborhood with parallel streets. If one street passes through (4, 5) and (3, 2), what is the equation for a parallel street that passes through (2, −3)?

Answers

Answer:

y=3x+(-9).

OR

y=3x-9

Step-by-step explanation:

First of all, we can find the slope of the first line.

m=[tex]\frac{y2-y1}{x2-x1}[/tex]

m=[tex]\frac{5-2}{4-3}[/tex]

m=3

We know that the parallel line will have the same slope as the first line. Now it's time to find the y-intercept of the second line.

To find the y-intercept, substitute in the values that we know for the second line.

(-3)=(3)(2)+b

(-3)=6+b

b=(-9)

Therefore, the final equation will be y=3x+(-9).

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Find the general solution of the differential equation y" - 81y = -243t + 162t². NOTE: Use t as the independent variable. Use c₁ and cg as arbitrary constants. C1 y(t) =

Answers

The general solution to the second order homogenous differential equation is  [tex]\(C_1 y(t) = c_1 e^{9t} + c_2 e^{-9t} - 2t^2 + 3t - \frac{4}{81}\)[/tex], where c₁ is a constant multiple of the entire expression.

What is the general solution to the differential equation?

To find the general solution of the given differential equation y'' - 81y = -243t + 162t², we can start by finding the complementary solution by solving the associated homogeneous equation y'' - 81y = 0.

The characteristic equation for the homogeneous equation is:

r² - 81 = 0

Factoring the equation:

(r - 9)(r + 9) = 0

This equation has two distinct roots: r = 9 and r = -9

Therefore, the complementary solution is:

[tex]\(y_c(t) = c_1 e^{9t} + c_2 e^{-9t}\)[/tex]    where c₁ and c₂ are arbitrary constants

To find a particular solution to the non-homogeneous equation, we can use the method of undetermined coefficients. Since the right-hand side of the equation is a polynomial in t of degree 2, we'll assume a particular solution of the form:

[tex]\(y_p(t) = At^2 + Bt + C\)[/tex]

Substituting this assumed form into the original differential equation, we can determine the values of A, B, and C. Taking the derivatives of [tex]\(y_p(t)\)[/tex]:

[tex]\(y_p'(t) = 2At + B\)\\\(y_p''(t) = 2A\)[/tex]

Plugging these derivatives back into the differential equation:

[tex]\(y_p'' - 81y_p = -243t + 162t^2\)\\\(2A - 81(At^2 + Bt + C) = -243t + 162t^2\)[/tex]

Simplifying the equation:

-81At² - 81Bt - 81C + 2A = -243t + 162t²

Now, equating the coefficients of the terms on both sides:

-81A = 162   (coefficients of t² terms)

-81B = -243  (coefficients of t terms)

-81C + 2A = 0  (constant terms)

From the first equation, we find A = -2.

From the second equation, we find B = 3.

Plugging these values into the third equation, we can solve for C:

-81C + 2(-2) = 0

-81C - 4 = 0

-81C = 4

C = -4/81

Therefore, the particular solution is:

[tex]\(y_p(t) = -2t^2 + 3t - \frac{4}{81}\)[/tex]

The general solution of the differential equation is the sum of the complementary and particular solutions:

[tex]\(y(t) = y_c(t) + y_p(t)\)\(y(t) = c_1 e^{9t} + c_2 e^{-9t} - 2t^2 + 3t - \frac{4}{81}\)[/tex]

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The general solution of the given differential equation is:

y(t) = c₁e^(9t) + c₂e^(-9t) - 2t² + 3t, where c₁ and c₂ are arbitrary constants.

To find the general solution of the given differential equation y" - 81y = -243t + 162t², we can solve it by first finding the complementary function and then a particular solution.

Complementary Function:

Let's find the complementary function by assuming a solution of the form y(t) = e^(rt).

Substituting this into the differential equation, we get:

r²e^(rt) - 81e^(rt) = 0

Factoring out e^(rt), we have:

e^(rt)(r² - 81) = 0

For a nontrivial solution, we require r² - 81 = 0. Solving this quadratic equation, we find two distinct roots: r = 9 and r = -9.

Therefore, the complementary function is given by:

y_c(t) = c₁e^(9t) + c₂e^(-9t), where c₁ and c₂ are arbitrary constants.

Particular Solution:

To find a particular solution, we can assume a polynomial of degree 2 for y(t) due to the right-hand side being a quadratic polynomial.

Let's assume y_p(t) = At² + Bt + C, where A, B, and C are constants to be determined.

Differentiating twice, we find:

y_p'(t) = 2At + B

y_p''(t) = 2A

Substituting these derivatives into the differential equation, we have:

2A - 81(At² + Bt + C) = -243t + 162t²

Comparing coefficients of like powers of t, we get the following equations:

-81A = 162 (coefficient of t²)

-81B = -243 (coefficient of t)

-81C + 2A = 0 (constant term)

Solving these equations, we find A = -2, B = 3, and C = 0.

Therefore, the particular solution is:

y_p(t) = -2t² + 3t

The general solution is the sum of the complementary function and the particular solution:

y(t) = y_c(t) + y_p(t)

= c₁e^(9t) + c₂e^(-9t) - 2t² + 3t

Therefore, the general solution of the given differential equation is:

y(t) = c₁e^(9t) + c₂e^(-9t) - 2t² + 3t, where c₁ and c₂ are arbitrary constants.

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(a) Find the solutions of the recurrence relation an ·an-1-12an-2 = 0, n ≥ 2, satisfying the initial conditions ao = 1,a₁ = 1
(b) Find the solutions of the recurrence relation a_n = 10a_(n-1) - 25a_(n-2) + 32, n ≥ 2, satisfying the initial conditions ao = 3, a₁ = 7. (c) Find all solutions of the recurrence relation a_n + a_(n-1) - 12a_(n-2) = 2^(n) (d) Find all the solutions of the recurrence relation a_n = 4a_(n-1) - 4a_(n-2)
(e) Find all the solutions of the recurrence relation a_n = 2a_(n-1) - a_(n-2) + 2
(f) Find all the solutions of the recurrence relation a_n - 2a_(n-1) - 3a_(n-2) = 3^(n)

Answers

Solutions for the given recurrence relations:

(a) Solutions for an ·an-1-12an-2 = 0, n ≥ 2, with ao = 1 and a₁ = 1.

(b) Solutions for a_n = 10a_(n-1) - 25a_(n-2) + 32, n ≥ 2, with ao = 3 and a₁ = 7.

(c) Solutions for a_n + a_(n-1) - 12a_(n-2) = 2^(n).

(d) Solutions for a_n = 4a_(n-1) - 4a_(n-2).

(e) Solutions for a_n = 2a_(n-1) - a_(n-2) + 2.

(f) Solutions for a_n - 2a_(n-1) - 3a_(n-2) = 3^(n).

In (a), the recurrence relation is an ·an-1-12an-2 = 0, and the initial conditions are ao = 1 and a₁ = 1. Solving this relation involves identifying the values of an that make the equation true.

In (b), the recurrence relation is a_n = 10a_(n-1) - 25a_(n-2) + 32, and the initial conditions are ao = 3 and a₁ = 7. Similar to (a), finding solutions involves identifying the values of a_n that satisfy the given relation.

In (c), the recurrence relation is a_n + a_(n-1) - 12a_(n-2) = 2^(n). Here, the task is to find all solutions of a_n that satisfy the relation for each value of n.

In (d), the recurrence relation is a_n = 4a_(n-1) - 4a_(n-2). Solving this relation entails determining the values of a_n that make the equation true.

In (e), the recurrence relation is a_n = 2a_(n-1) - a_(n-2) + 2. The goal is to find all solutions of a_n that satisfy the relation for each value of n.

In (f), the recurrence relation is a_n - 2a_(n-1) - 3a_(n-2) = 3^(n). Solving this relation involves finding all values of a_n that satisfy the equation.

Solving recurrence relations is an essential task in understanding the behavior and patterns within a sequence of numbers. It requires analyzing the relationship between terms and finding a general expression or formula that describes the sequence. By utilizing the given initial conditions, the solutions to the recurrence relations can be determined, providing insights into the values of the sequence at different positions.

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A 9th order, linear, homogeneous, constant coefficient differential equation has a characteristic equation which factors as follows. (r² − 4r+8)³√(r + 2)² = 0 Write the nine fundamental solutions to the differential equation. y₁ = Y4= Y1 = y₂ = Y5 = Y8 = Уз = Y6 = Y9 =

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The fundamental solutions to the differential equation are:

y1 = e^(2x)sin(2x)y2 = e^(2x)cos(2x)y3 = e^(-2x)y4 = xe^(2x)sin(2x)y5 = xe^(2x)cos(2x)y6 = e^(2x)sin(2x)cos(2x)y7 = xe^(-2x)y8 = x²e^(2x)sin(2x)y9 = x²e^(2x)cos(2x)

The characteristic equation that factors in a 9th order, linear, homogeneous, constant coefficient differential equation is (r² − 4r+8)³√(r + 2)² = 0.

To solve this equation, we need to split it into its individual factors.The factors are: (r² − 4r+8)³ and (r + 2)²

To determine the roots of the equation, we'll first solve the quadratic equation that represents the first factor: (r² − 4r+8) = 0.

Using the quadratic formula, we get:

r = (4±√(16−4×1×8))/2r = 2±2ir = 2+2i, 2-2i

These are the complex roots of the quadratic equation. Because the root (r+2) has a power of two, it has a total of four roots:r = -2, -2 (repeated)

Subsequently, the total number of roots of the characteristic equation is 6 real roots (two from the quadratic equation and four from (r+2)²) and 6 complex roots (three from the quadratic equation)

Because the roots are distinct, the nine fundamental solutions can be expressed in terms of each root. Therefore, the fundamental solutions to the differential equation are:

y1 = e^(2x)sin(2x)

y2 = e^(2x)cos(2x)

y3 = e^(-2x)y4 = xe^(2x)sin(2x)

y5 = xe^(2x)cos(2x)

y6 = e^(2x)sin(2x)cos(2x)

y7 = xe^(-2x)

y8 = x²e^(2x)sin(2x)

y9 = x²e^(2x)cos(2x)

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help if you can asap pls an thank you!!!!

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Answer: SSS

Step-by-step explanation:

The lines on the triangles say that 2 of the sides are equal. Th triangles also share a 3rd side that is equal.

So, a side, a side and a side proves the triangles are congruent through, SSS

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