The magnitudes of vectors u and v and the angle
θ
between the vectors are given. Find the sum of
u+v.
​|u​|=24​,
​|v​|=24​,
θ=129

Given,Face value (FV) of the note = $1000Issued date = April 21, 2005Rate of interest (r) = 5.5%Time period (t) = 90 daysNow, we have to find the maturity value of the note.To compute the maturity value, we have to find the interest and then add it to the face value (FV) of the note.

To find the interest, we use the formula,Interest (I) = (FV x r x t) / (100 x 365)where t is in days.Putting the given values in the above formula, we get,I = (1000 x 5.5 x 90) / (100 x 365)= 150.14So, the interest on the note is $150.14.Now, the maturity value (MV) of the note is given by,MV = FV + I= $1000 + $150.14= $1150.14Therefore, the maturity value of the note is $1150.14.

On computing the maturity value of a 90-day note with a face value of $1000 issued on April 21, 2005, at an interest rate of 5.5%, it is found that the maturity value of the note is $1150.14.

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The given differential equation is not exact. We can use the definition of an exact differential equation to determine whether the given differential equation is exact or not.

An equation of the form M(x, y)dx + N(x, y)dy = 0 is called exact if and only if there exists a function Φ(x, y) such that the total differential of Φ(x, y) is given by dΦ = ∂Φ/∂xdx + ∂Φ/∂ydy anddΦ = M(x, y)dx + N(x, y)dy.On comparing the coefficients of dx, we get ∂M/∂y = 0and on comparing the coefficients of dy, we get ∂N/∂x = 0.Here, we have M(x, y) = 2x - 1 and N(x, y) = 5y + 8∂M/∂y = 0, but ∂N/∂x = 0 is not true. Therefore, the given differential equation is not exact. The answer is NOT.

Now, we can use an integrating factor to solve the differential equation. An integrating factor, μ(x, y) is a function which when multiplied to the given differential equation, makes it exact. The general formula for an integrating factor is given by:μ(x, y) = e^(∫(∂N/∂x - ∂M/∂y) dy)Here, ∂N/∂x - ∂M/∂y = 5 - 0 = 5.We have to multiply the given differential equation by μ(x, y) = e^(∫(∂N/∂x - ∂M/∂y) dy) = e^(5y)and get an exact differential equation.(2x - 1)e^(5y)dx + (5y + 8)e^(5y)dy = 0We now have to find the function Φ(x, y) such that its total differential is the given equation.Let Φ(x, y) be a function such that ∂Φ/∂x = (2x - 1)e^(5y) and ∂Φ/∂y = (5y + 8)e^(5y).

Integrating ∂Φ/∂x w.r.t x, we get:Φ(x, y) = ∫(2x - 1)e^(5y) dx Integrating ∂Φ/∂y w.r.t y, we get:Φ(x, y) = ∫(5y + 8)e^(5y) dySo, we have:∫(2x - 1)e^(5y) dx = ∫(5y + 8)e^(5y) dy Differentiating the first expression w.r.t y and the second expression w.r.t x, we get:(∂Φ/∂y)(∂y/∂x) = (2x - 1)e^(5y)and (∂Φ/∂x)(∂x/∂y) = (5y + 8)e^(5y) Comparing the coefficients of e^(5y), we get:∂Φ/∂y = (2x - 1)e^(5y) and ∂Φ/∂x = (5y + 8)e^(5y)

Therefore, the solution to the differential equation is given by:Φ(x, y) = ∫(2x - 1)e^(5y) dx = (x^2 - x)e^(5y) + Cwhere C is a constant. Thus, the solution to the given differential equation is given by:(x^2 - x)e^(5y) + C = 0

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The projected revenue from the sale of unit 46 would be $142,508.

To find the marginal revenue, we first take the derivative of the revenue function R(x):

R'(x) = d/dx(66x² + 73x + 2x + 2)

R'(x) = 132x + 73 + 2

Next, we substitute x = 45 into the marginal revenue function:

R'(45) = 132(45) + 73 + 2

R'(45) = 5940 + 73 + 2

R'(45) = 6015

Therefore, the marginal revenue when 45 units are sold is $6,015.

To estimate the projected revenue from the sale of unit 46, we evaluate the revenue function at x = 46:

R(46) = 66(46)² + 73(46) + 2(46) + 2

R(46) = 66(2116) + 73(46) + 92 + 2

R(46) = 139,056 + 3,358 + 92 + 2

R(46) = 142,508

Hence, the projected revenue from the sale of unit 46 would be $142,508.

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The exact solutions of the equation \(\cos(3x) = -\frac{1}{\sqrt{2}}\) in the interval \([0, \pi)\) are \(x = \frac{\pi}{6}, \frac{5\pi}{6}\).

To find the solutions, we can start by determining the angles whose cosine is \(-\frac{1}{\sqrt{2}}\). Since the cosine function is negative in the second and third quadrants, we need to find the angles in those quadrants whose cosine is \(\frac{1}{\sqrt{2}}\).
In the second quadrant, the reference angle with cosine \(\frac{1}{\sqrt{2}}\) is \(\frac{\pi}{4}\). Therefore, one solution is \(x = \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4}\).
In the third quadrant, the reference angle with cosine \(\frac{1}{\sqrt{2}}\) is also \(\frac{\pi}{4}\). Therefore, another solution is \(x = \pi - \frac{\pi}{4} = \frac{3\pi}{4}\).
Since we are looking for solutions in the interval \([0, \pi)\), we only consider the solutions that lie within this range. Therefore, the exact solutions in the given interval are \(x = \frac{\pi}{6}, \frac{5\pi}{6}\).
Hence, the solutions to the equation \(\cos(3x) = -\frac{1}{\sqrt{2}}\) in the interval \([0, \pi)\) are \(x = \frac{\pi}{6}, \frac{5\pi}{6}\).



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To find the compositions of f(x) = 2x and g(x) = x given in the problem, that is fog, gof, and 9°9, we first need to understand what each of them means. Composition of functions is an operation that takes two functions f(x) and g(x) and creates a new function h(x) such that h(x) = f(g(x)).

For example, if f(x) = 2x and g(x) = x + 1, then their composition, h(x) = f(g(x)) = 2(x + 1) = 2x + 2. Here, we have f(x) = 2x and g(x) = x.(a) fog We can find fog as follows: fog(x) = f(g(x)) = f(x) = 2x

Therefore, fog(x) = 2x.(b) gofWe can find gof as follows: gof(x) = g(f(x)) = g(2x) = 2x

Therefore, gof(x) = 2x.(c) 9°9We cannot find 9°9 because it is not a valid composition of functions

. The symbol ° is typically used to denote composition, but in this case, it is unclear what the functions are that are being composed.

Therefore, we cannot find 9°9. We have found that fog(x) = 2x and gof(x) = 2x.

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The reason number 3 include the following: D. Definition of midpoint.

In Mathematics and Geometry, a midpoint is a point that lies exactly at the middle of two other end points that are located on a straight line segment.

In this context, we can prove that line segment AC is congruent to line segment BC by completing the two-column proof shown above with the following reasons from step 1 to step 3:

Statements                                Reasons

1. M is the midpoint of AB           Given

2. AB ⊥CM                                   Given

3. AM ≅ BM                             Definition of midpoint

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The expansion of the binomial [tex](x+y)^2a+5[/tex] has 20 terms. the value of a

is 7.

To determine the value of "a" in the expansion of the binomial [tex](x+y)^(2a+5)[/tex] with 20 terms, we need to use the concept of binomial expansion and the formula for the number of terms in a binomial expansion.

The formula for the number of terms in a binomial expansion is given by (n + 1), where "n" represents the power of the binomial. In this case, the power of the binomial is (2a + 5). Therefore, we have:

(2a + 5) + 1 = 20

Simplifying the equation:

2a + 6 = 20

Subtracting 6 from both sides:

2a = 20 - 6

2a = 14

Dividing both sides by 2:

a = 14 / 2

a = 7

Therefore, the value of "a" is 7.

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Looking at the graph, the correct answer is in option B; An employee would earn $730 if they work for 27 hours on a project.

A scatterplot is a type of graphical representation that displays the relationship between two numerical variables. It is particularly useful for visualizing the correlation or pattern between two sets of data points.

We can see that we can trace the statement that is correct when we try to match each of the points on the graph. When we do that, we can see that 27 hours can be matched with $730 earnings.

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The range of the function \( f(x) = \csc(x) \) is the set of all real numbers except for \( -1 \) and \( 1 \). The range of the function \( f(x) = \arcsin(x) \) is \([- \frac{\pi}{2}, \frac{\pi}{2}]\).

For the function \( f(x) = \cos(x) \), the range represents the set of all possible values that \( f(x) \) can take. Since the cosine function oscillates between \( -1 \) and \( 1 \) for all real values of \( x \), the range is \([-1, 1]\).

In the case of \( f(x) = \csc(x) \), the range is the set of all real numbers except for \( -1 \) and \( 1 \). The cosecant function is defined as the reciprocal of the sine function, and it takes on all real values except for the points where the sine function crosses the x-axis (i.e., \( -1 \) and \( 1 \)).

Finally, for \( f(x) = \arcsin(x) \), the range represents the set of all possible outputs of the inverse sine function. Since the domain of the inverse sine function is \([-1, 1]\), the range is \([- \frac{\pi}{2}, \frac{\pi}{2}]\) in radians, which corresponds to \([-90^\circ, 90^\circ]\) in degrees.

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The rocket peaks at 906.43 meters above sea-level.

Given: h(t)=-4.9t² + 139t + 346

We know that the rocket will splash down into the ocean means the height of the rocket at splashdown will be 0,

So let's solve the first part of the question to find the time at which splashdown occur.

h(t)=-4.9t² + 139t + 346

Putting h(t) = 0,-4.9t² + 139t + 346 = 0

Multiplying by -10 on both sides,4.9t² - 139t - 346 = 0

Solving the above quadratic equation, we get, t = 28.7 s (approximately)

The rocket will splash down after 28.7 seconds.

Now, to find the height at the peak, we can use the formula t = -b / 2a,

which gives us the time at which the rocket reaches the peak of its flight.

h(t) = -4.9t² + 139t + 346

Differentiating w.r.t t, we get dh/dt = -9.8t + 139

Putting dh/dt = 0 to find the maximum height-9.8t + 139 = 0t = 14.18 s (approximately)

So, the rocket reaches the peak at 14.18 seconds

The height at the peak can be found by putting t = 14.18s in the equation

h(t)=-4.9t² + 139t + 346

h(14.18) = -4.9(14.18)² + 139(14.18) + 346 = 906.43 m

The rocket peaks at 906.43 meters above sea-level.

To find the time at which splashdown occur, we need to put h(t) = 0 in the given function of the height of the rocket, and solve the quadratic equation that results.

The time at which the rocket reaches the peak can be found by calculating the time at which the rate of change of height is 0 (i.e., when the derivative of the height function is 0).

We can then find the height at the peak by plugging in this time into the original height function.

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The solution to the differential equation is [tex]$y = \left(\frac{7}{2}\left(-\frac{1}{6}y^{\frac{2}{7}} e^{-6x} - \frac{1}{36}e^{-6x}y^{\frac{6}{7}} + \frac{2}{7}\right)\right)^{\frac{1}{5}}$[/tex]

Given DE : [tex]$y^{\frac{1}{7}} \frac{dy}{dx} + y^{\frac{3}{2}} = 1$[/tex] and the initial value y(0) = 0

This is a Bernoulli differential equation. It can be converted to a linear differential equation by substituting[tex]$v = y^{1-7}$[/tex], we get [tex]$\frac{dv}{dx} + (1-7)v = 1- y^{-\frac{1}{2}}$[/tex]

On simplification, [tex]$\frac{dv}{dx} - 6v = y^{-\frac{1}{2}}$[/tex]

The integrating factor [tex]$I = e^{\int -6 dx} = e^{-6x}$On[/tex] multiplying both sides of the equation by I, we get

[tex]$I\frac{dv}{dx} - 6Iv = y^{-\frac{1}{2}}e^{-6x}$[/tex]

Rewriting the LHS,

[tex]$\frac{d}{dx} (Iv) = y^{-\frac{1}{2}}e^{-6x}$[/tex]

On integrating both sides, we get

[tex]$Iv = \int y^{-\frac{1}{2}}e^{-6x}dx + C_1$[/tex]

On substituting back for v, we get

[tex]$y^{1-7} = \int y^{-\frac{1}{2}}e^{-6x}dx + C_1e^{6x}$[/tex]

On simplification, we get

[tex]$y = \left(\int y^{\frac{5}{7}}e^{-6x}dx + C_1e^{6x}\right)^{\frac{1}{5}}$[/tex]

On integrating, we get

[tex]$I = \int y^{\frac{5}{7}}e^{-6x}dx$[/tex]

For finding I, we can use integration by substitution by letting

[tex]$t = y^{\frac{2}{7}}$ and $dt = \frac{2}{7}y^{-\frac{5}{7}}dy$.[/tex]

Then [tex]$I = \frac{7}{2} \int e^{-6x}t dt = \frac{7}{2}\left(-\frac{1}{6}t e^{-6x} - \frac{1}{36}e^{-6x}t^3 + C_2\right)$[/tex]

On substituting [tex]$t = y^{\frac{2}{7}}$, we get$I = \frac{7}{2}\left(-\frac{1}{6}y^{\frac{2}{7}} e^{-6x} - \frac{1}{36}e^{-6x}y^{\frac{6}{7}} + C_2\right)$[/tex]

Finally, substituting for I in the solution, we get the general solution

[tex]$y = \left(\frac{7}{2}\left(-\frac{1}{6}y^{\frac{2}{7}} e^{-6x} - \frac{1}{36}e^{-6x}y^{\frac{6}{7}} + C_2\right) + C_1e^{6x}\right)^{\frac{1}{5}}$[/tex]

On applying the initial condition [tex]$y(0) = 0$[/tex], we get[tex]$C_1 = 0$[/tex]

On applying the initial condition [tex]$y(0) = 0$, we get$C_2 = \frac{2}{7}$[/tex]

So the solution to the differential equation is

[tex]$y = \left(\frac{7}{2}\left(-\frac{1}{6}y^{\frac{2}{7}} e^{-6x} - \frac{1}{36}e^{-6x}y^{\frac{6}{7}} + \frac{2}{7}\right)\right)^{\frac{1}{5}}$[/tex]

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The gcd and lcm of the pairs of integers are as follows:

(a) For \(a = 756\) and \(b = 210\), the gcd is 42 and the lcm is 3780.

(b) For \(a = 346\) and \(b = 874\), the gcd is 2 and the lcm is 60148.

In the first pair of integers, 756 and 210, we can apply the Euclidean algorithm to find the gcd. We divide 756 by 210, which gives us a quotient of 3 and a remainder of 126. Next, we divide 210 by 126, resulting in a quotient of 1 and a remainder of 84. Continuing this process, we divide 126 by 84, obtaining a quotient of 1 and a remainder of 42. Finally, we divide 84 by 42, and the remainder is 0. Therefore, the gcd is the last non-zero remainder, which is 42. To find the lcm, we use the formula lcm(a, b) = (a * b) / gcd(a, b). Plugging in the values, we get lcm(756, 210) = (756 * 210) / 42 = 3780.

In the second pair of integers, 346 and 874, we repeat the same steps. We divide 874 by 346, resulting in a quotient of 2 and a remainder of 182. Next, we divide 346 by 182, obtaining a quotient of 1 and a remainder of 164. Continuing this process, we divide 182 by 164, and the remainder is 18. Finally, we divide 164 by 18, and the remainder is 2. Therefore, the gcd is 2. To find the lcm, we use the formula lcm(a, b) = (a * b) / gcd(a, b). Plugging in the values, we get lcm(346, 874) = (346 * 874) / 2 = 60148.

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The given differential equation is y'' + (x - 2)y' + y = 0. It can be solved using power series expansion at x = 0 for a general solution to the given differential equation.

To find the power series expansion of the solution of the given differential equation, we can use the following steps:

Step 1: Let y(x) = Σ an xⁿ.

Step 2: Substitute y and its derivatives in the differential equation: y'' + (x - 2)y' + y = 0.

            After simplifying, we get:

            => [Σ n(n-1)an xⁿ-2] + [Σ n(n-1)an xⁿ-1] - [2Σ n an xⁿ-1] + [Σ an xⁿ] = 0.

Step 3: For this equation to hold true for all values of x, all the coefficients of the like powers of x should be zero.                                              

            Hence, we get the following recurrence relation:

            => (n+2)(n+1)an+2 + (2-n)an = 0.

Step 4: Solve the recurrence relation to find the values of the coefficients an.

            => an+2 = - (2-n)/(n+2) * an.

Step 5: Therefore, the solution of the differential equation is given by:

             => y(x) = Σ an xⁿ = a0 + a1 x + a2 x² + a3 x³ + ...

                  where, a0, a1, a2, a3, ... are arbitrary constants.

Step 6: Now we need to find the first four non-zero terms of the power series expansion of y(x) about x = 0.

            We know that at x = 0, y(x) = a0.

            Using the recurrence relation, we can write the value of a2 in terms of a0 as:

            => a2 = -1/2 * a0

            Using the recurrence relation again, we can write the value of a3 in terms of a0 and a2 as:

            => a3 = 1/3 * a2 = -1/6 * a0

Step 7: Therefore, the first four nonzero terms in a power series expansion about x = 0 for a general solution to the given differential equation are given by the below expression:

            y(x) = a0 - 1/2 * a0 x² - 1/6 * a0 x³ + 1/24 * a0 x⁴.

Hence, the answer is y(x) = a0 - 1/2 * a0 x² - 1/6 * a0 x³ + 1/24 * a0 x⁴

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The domain of H(t) is given as follows:

B. 0 ≤ t ≤ 36.

The values of x of the graph range from 0 to 36, hence the domain of the function is given as follows:

B. 0 ≤ t ≤ 36.

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The exponentiation is associative, and the equation `(a^b)^c=a^(b*c)` is correct for all integers.

Suppose there are two integers, `a` and `b`. define the operation "^" as follows: ^= This operation is also known as exponentiation. find out if exponentiation is associative. The following is always true:

`(a^b)^c

=a^(b*c)`

Assume `a=2, b=3,` and `c=4`.

Let's use the above formula to find the left-hand side of the equation:

`(2^3)^4

=8^4

=4096`

Using the same values of `a`, `b`, and `c`, use the formula to calculate the right-hand side of the equation: `2^(3*4)

=2^12

=4096`

The answer to both sides is `4096`, indicating that exponentiation is associative, and the equation `(a^b)^c=a^(b*c)` is correct for all integers.

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We have shown that y = (441 + x^2) / 42 based on the properties of similar triangles.

To determine the value of y in terms of x, we will use the properties of similar triangles.

In triangle CDE, we can see that triangle CDE is similar to triangle CAB. This is because angle CDE and angle CAB are both right angles, and angle CED and angle CAB are congruent due to the folding process.

Let's denote the length of AC as y cm and ED as x cm.

Since triangle CDE is similar to triangle CAB, we can set up the following proportion:

CD/AC = CE/AB

CD is equal to the length of the A4-size paper, which is 29.7 cm, and AB is the width of the paper, which is 21 cm.

So we have:

29.7/y = x/21

Cross-multiplying:

29.7 * 21 = y * x

623.7 = y * x

Dividing both sides of the equation by y:

623.7/y = y * x / y

623.7/y = x

Now, to express y in terms of x, we rearrange the equation:

y = 623.7 / x

Simplifying further:

y = (441 + 182.7) / x

y = (441 + x^2) / x

y = (441 + x^2) / 42

Therefore, we have shown that y = (441 + x^2) / 42 based on the properties of similar triangles.

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Naruto paid an interest of $55.25 to his credit card company for the purchase of his TV.

The interest Naruto paid for the purchase of his TV can be calculated using the simple interest formula:

Interest = Principal × Rate × Time

In this case, the principal is $850, the rate is 13% (or 0.13 as a decimal), and the time is 6 months (or 0.5 years). Plugging these values into the formula, we get:

Interest = $850 × 0.13 × 0.5 = $55.25

Therefore, Naruto paid an interest of $55.25 to his credit card company for the purchase of his TV.

The correct answer is option d. Naruto paid an interest of $55.25.

It's important to note that in this scenario, Naruto paid off the total cost of the TV after six months. Since the interest rate is annual, the interest is calculated based on the principal amount for the duration of six months. If Naruto had taken longer to pay off the TV or had not paid it off within a year, the interest amount would have been higher. However, in this case, Naruto paid off the TV before a year's time, so the interest amount is relatively low.

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The correct answers are:

d)The median is the only measurement that must always be one of the data values in your set of data.

a)Mean = 357n ; Median = 3629 & Mode = 3629

b)Shortest height: 2722 Tallest height: 4791

c)Range = 2069

d)The standard-deviation of the height data is 384.44.

d. If your data set has a mean, median, and mode, the median is the only measurement that must always be one of the data values in your set of data.

This is because the median is the middle value in a data set, so it must be one of the actual data values in order to represent the center of the distribution.

The mean and mode, on the other hand, can be influenced by outliers or skewed data, so they do not necessarily have to be actual data values in the set.

Therefore, the median is the measurement that always represents a true value in the data set.
Given that the height data statistics are:
a. Mean = 357n
Median = 3629
Mode = 3629
b. The shortest and tallest height values are:
Shortest: 2722
Tallest: 4791
c. The range of the data is:
Range = Tallest height – Shortest height
Range = 4791 – 2722
Range = 2069
d. To calculate the standard deviation of the height data:
Using Excel, the standard deviation formula is :
STDEV.P(data range), which gives a result of 384.44.
Therefore, the standard deviation of the height data is 384.44.

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The probability that either Event A and Event B occur can be determined by calculating the sum of their individual probabilities minus the probability that both events occur simultaneously.

Let's find the probability that Event A occurs first: P(A) = 3/19Next, let's determine the probability that Event B occurs: P(B) = 2/19The probability that both Event A and Event B occur simultaneously can be found as follows: P(A and B) = Middle 10/19Therefore, the probability that either.

Event A or Event B occur can be calculated using the following formula: P(A or B) = P(A) + P(B) - P(A and B)Substituting the values from above, we get:P(A or B) = 3/19 + 2/19 - 10/19P(A or B) = -5/19However, this result is impossible since probabilities are always positive. Hence, there has been an error in the data provided.

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The roots of the quadratic equation obtained using the quadratic formula method are [tex]$\frac{4}{3}$ and $\frac{5}{3}$.[/tex]

The method used to factorize the expression -3x² + 8x-5 is completing the square method.

That coefficient is half of the coefficient of the x term squared; in this case, it is (8/(-6))^2 = (4/3)^2 = 16/9.

So, we have -3x² + 8x - 5= -3(x^2 - 8x/3 + 16/9 - 5 - 16/9)= -3[(x - 4/3)^2 - 49/9]

By simplifying the above expression, we get the final answer which is: -3(x - 4/3 + 7/3)(x - 4/3 - 7/3)

Now, we can use another method of factorization to check the answer is correct.

Let's use the quadratic formula.

The quadratic formula is given by:

                    [tex]$$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$[/tex]

Comparing with our expression, we get a=-3, b=8, c=-5

Putting these values in the quadratic formula and solving it, we get

        [tex]$x=\frac{-8\pm \sqrt{8^2 - 4(-3)(-5)}}{2(-3)}$[/tex]

which simplifies to:

              [tex]$x=\frac{4}{3} \text{ or } x=\frac{5}{3}$[/tex]

Hence, the factors of the given expression are [tex]$(x - 4/3 + 7/3)(x - 4/3 - 7/3)$.[/tex]

The roots of the quadratic equation obtained using the quadratic formula method are [tex]$\frac{4}{3}$ and $\frac{5}{3}$.[/tex]

As we can see, both methods of factorisation gave the same factors, which proves that the answer is correct.

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The simplified form of the given trigonometric expressions are (sinθ + tanθ)/cos²θ.

Given expressions are

sinθ/(1 - secθ) and (1 + secθ)/(1 - secθ)

To simplify the expressions, we can multiply the numerators and the denominators together,

sinθ × (1 + secθ)/(1 - secθ) × (1 + secθ)

Now simplify the numerator

sinθ × (1 + secθ) = sinθ + sinθ × secθ

Now simplify the denominator

(1 - secθ) × (1 + secθ) = (1 - sec²θ)

We can use the identity (1 - sec²θ) = cos²θ to rewrite the denominator

(1 - secθ) × (1 + secθ) = cos²θ

Putting the simplified numerator and denominator back together, we have

= (sinθ + sinθsecθ)/cos²θ

We can simplify this expression further. Let's factor out a common factor of sinθ from the numerator

= sinθ(1 + secθ)/cos²θ

Use the identity secθ = 1/cosθ, rewrite the numerator as

= sinθ(1 + 1/cosθ)/cos²θ

= (sinθ + sinθ/cosθ)/cos²θ

Use the identity sinθ/cosθ = tanθ

= (sinθ + tanθ)/cos²θ

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A reference number is a real number ranging from -1 to 1, representing the angle created when a point is placed on the terminal side of an angle in the standard position. It can be calculated using trigonometric functions sine, cosine, and tangent. For t values of 4π/7, -7π/9, -3, and 5, the reference numbers are 0.50 + 0.86i, -0.62 + 0.78i, -0.99 + 0.14i, and 0.28 - 0.96i.

A reference number is a real number that ranges from -1 to 1. It represents the angle created when a point is placed on the terminal side of an angle in the standard position. The trigonometric functions sine, cosine, and tangent can be used to calculate the reference number.

Let's consider the given values of t. (a) t=47π4(a) We know that the reference angle θ is given by 

θ = |t| mod 2π.θ

= (4π/7) mod 2π

= 4π/7

Therefore, the reference angle θ is 4π/7. Now, we can calculate the value of sinθ and cosθ which represent the reference number. sin(4π/7) = 0.86 (approx)cos(4π/7) = 0.50 (approx)Thus, the reference number for t = 4π/7 is cos(4π/7) + i sin(4π/7)

= 0.50 + 0.86i.

(b) t=-79(a) We know that the reference angle θ is given by θ = |t| mod 2π.θ = (7π/9) mod 2π= 7π/9Therefore, the reference angle θ is 7π/9. Now, we can calculate the value of sinθ and cosθ which represent the reference number.sin(7π/9) = 0.78 (approx)cos(7π/9) = -0.62 (approx)Thus, the reference number for

t = -7π/9 is cos(7π/9) + i sin(7π/9)

= -0.62 + 0.78i. (c)

t=-3(b) 

We know that the reference angle θ is given by

θ = |t| mod 2π.θ

= 3 mod 2π

= 3

Therefore, the reference angle θ is 3. Now, we can calculate the value of sinθ and cosθ which represent the reference number.sin(3) = 0.14 (approx)cos(3) = -0.99 (approx)Thus, the reference number for t = -3 is cos(3) + i sin(3) = -0.99 + 0.14i. (d) t=5(c) We know that the reference angle θ is given by θ = |t| mod 2π.θ = 5 mod 2π= 5Therefore, the reference angle θ is 5.

Now, we can calculate the value of sinθ and cosθ which represent the reference number.sin(5) = -0.96 (approx)cos(5) = 0.28 (approx)Thus, the reference number for t = 5 is cos(5) + i sin(5)

= 0.28 - 0.96i. Thus, the reference numbers for the given values of t are (a) 0.50 + 0.86i, (b) -0.62 + 0.78i, (c) -0.99 + 0.14i, and (d) 0.28 - 0.96i.

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a) The solutions to f(x) = 0 are x = 1 and x = -1.

b)   the solution to g(x) = 0 is x = -1/5.

C)   the right-hand side of this equation is negative for all real values of x, there are no real solutions to f(x) = g(x).

d)   the solution to f(x) > 0 is (-∞,0) U (0,∞).

e)  We get: f(g(x)) = -25x² - 10x

g)   Interval notation, the solution to f(x) > 1 is (-√2,0) U (0,√2).

(a) To solve f(x) = 0, we substitute 0 for f(x) and solve for x:

-f(x)² + 1 = 0

-f(x)² = -1

f(x)² = 1

Taking the square root of both sides, we get:

f(x) = ±1

Therefore, the solutions to f(x) = 0 are x = 1 and x = -1.

(b) To solve g(x) = 0, we substitute 0 for g(x) and solve for x:

5x + 1 = 0

Solving for x, we get:

x = -1/5

Therefore, the solution to g(x) = 0 is x = -1/5.

(c) To solve f(x) = g(x), we substitute the expressions for f(x) and g(x) and solve for x:

-f(x)² + 1 = 5x + 1

Simplifying, we get:

-f(x)² = 5x

Dividing by -1, we get:

f(x)² = -5x

Since the right-hand side of this equation is negative for all real values of x, there are no real solutions to f(x) = g(x).

(d) To solve f(x) > 0, we look for the values of x that make f(x) positive. Since f(x) = -x² + 1, we know that f(x) is a downward-facing parabola with its vertex at (0,1). Therefore, f(x) is positive for all values of x that lie within the interval (-∞,0) or (0,∞). In interval notation, the solution to f(x) > 0 is (-∞,0) U (0,∞).

(e) To solve g(x) ≤ 0, we look for the values of x that make g(x) less than or equal to zero. Since g(x) = 5x + 1, we know that g(x) is a linear function with a positive slope of 5. Therefore, g(x) is less than or equal to zero for all values of x that lie within the interval (-∞,-1/5]. In interval notation, the solution to g(x) ≤ 0 is (-∞,-1/5].

(f) To solve f(g(x)), we substitute the expression for g(x) into f(x):

f(g(x)) = -g(x)² + 1

Substituting the expression for g(x), we get:

f(g(x)) = - (5x + 1)² + 1

Expanding and simplifying, we get:

f(g(x)) = -25x² - 10x

(g) To solve f(x) > 1, we look for the values of x that make f(x) greater than 1. Since f(x) = -x² + 1, we know that f(x) is a downward-facing parabola with its vertex at (0,1). Therefore, f(x) is greater than 1 for all values of x that lie within the intervals (-√2,0) or (0,√2). In interval notation, the solution to f(x) > 1 is (-√2,0) U (0,√2).

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Answer:

Step-by-step explanation:

\begin{align*}

T_{1,1} &= \frac{1}{2} (f(0) + f(1)) \\

&= \frac{1}{2} (1 + \frac{1}{2}) \\

&= \frac{3}{4}

\end{align*}

Now, for two subintervals:

\begin{align*}

T_{2,1} &= \frac{1}{4} (f(0) + 2f(1/2) + f(1)) \\

&= \frac{1}{4} \left(1 + 2 \left(\frac{1}{1 + \left(\frac{1}{2}\right)^2}\right) + \frac{1}{1^2}\right) \\

&= \frac{1}{4} \left(1 + 2 \left(\frac{1}{1 + \frac{1}{4}}\right) + 1\right) \\

&= \frac{1}{4} \left(1 + 2 \left(\frac{1}{\frac{5}{4}}\right) + 1\right) \\

&= \frac{1}{4} \left(1 + 2 \cdot \frac{4}{5} + 1\right) \\

&= \frac{1}{4} \left(1 + \frac{8}{5} + 1\right) \\

&= \frac{1}{4} \left(\frac{5}{5} + \frac{8}{5} + \frac{5}{5}\right)

\end{align*}

Thus, the approximate value of the integral using Romberg's method is T_2,1, and this can also be used to obtain an approximate value of π.

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By combining the approaches, researchers can gather comprehensive data, analyze existing information, conduct controlled experiments, and obtain direct responses through surveys.

Existing Data Analysis: Begin by collecting relevant existing data from reliable sources, such as research studies, government databases, or publicly available datasets. Identify variables related to the research question and extract the necessary data for analysis. Use statistical tools and techniques to examine the relationship between the IV and DV based on the existing data.

Experimentation: Design and conduct experiments to measure the IV and its impact on the DV. Clearly define the experimental conditions and variables, including the manipulation of the IV and the measurement of the resulting changes in the DV. Ensure appropriate control groups and randomization to minimize biases and confounding factors.

Survey Research: Develop a survey questionnaire to gather data directly from participants. Formulate specific questions that capture the IV and DV variables. Include options or response choices that cover a range of possibilities for the IV and capture the variations in the DV. Ensure the survey questions are clear, unbiased, and appropriately structured to elicit relevant responses.

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The total price of the parallelogram-shaped plot of land is approximately $4,884, given its area of 88,779 square units and a price of $2400 per acre.

To calculate the area of the parallelogram-shaped plot of land, we can use the formula:

Area = base length * height

Given the base lengths of 303 and 315 units and a height of 293 units, we can substitute these values into the formula:

Area = 303 * 293

Area = 88,779 square units

Now, to convert the area from square units to acres, we divide it by the conversion factor:

Area (in acres) = 88,779 / 43,560

Area (in acres) ≈ 2.035 acres

Finally, to find the total price of the land, we multiply the area in acres by the price per acre, which is $2400:

Total Price = 2.035 acres * $2400/acre

Total Price ≈ $4,884

Therefore, the total price of the land is approximately $4,884.

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The complete question is:

The parallelogram shaped plot of land shown in the figure to the right is put up for sale at $2400 per acre. What is the total price of the land?given that it has side lengths of 303 units and 315 units, a height of 293 units?

To create a line graph in Excel 2016 and display data points as dots, enter the data in two columns, select the data range, insert a line graph, add data series for each column, and customize the graph. Right-click on the lines, format data series, and choose marker options to display dots.

to create a line graph in Excel 2016 using the given data. Here's what you need to do:

Step 1: Open Excel and enter the data into two columns. Place the "X" values in column A (Points) and the "Y" values in column B (Screens and Shoes).

Step 2: Select the data range by clicking and dragging to highlight both columns.

Step 3: Go to the "Insert" tab in the Excel menu.

Step 4: In the "Charts" section, click on the "Line" button. Select the first line graph option from the drop-down menu.

Step 5: A basic line graph will be inserted onto your worksheet.

Step 6: Right-click on the graph and select "Select Data" from the context menu.

Step 7: In the "Select Data Source" dialog box, click the "Add" button under "Legend Entries (Series)."

Step 8: In the "Edit Series" dialog box, enter "Points" for the series name, select the data range for the X values (A2:A7), and select the data range for the Y values (B2:B7). Click "OK."

Step 9: Repeat steps 7 and 8 for the second series. Enter "Screens" for the series name, select the data range for the X values (A2:A7), and select the data range for the Y values (B2:B7). Click "OK."

Step 10: Your line graph will now display both series. You can customize the graph by adding titles, labels, and adjusting the formatting as desired.

To add data points as dots, follow these additional steps:

Step 11: Right-click on one of the lines in the graph and select "Format Data Series" from the context menu.

Step 12: In the "Format Data Series" pane, under "Marker Options," select the marker type you prefer, such as "Circle" or "Dot."

Step 13: Adjust the size and fill color of the markers, if desired.

Step 14: Click "Close" to apply the changes.

Your line graph with data points as dots should now be ready.

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Answer: x = 8

Step-by-step explanation:

The two lines are of the same length. We can write the equation 11 + 7x = 67 to represent this. We can simplify (solve) this equation by isolating our variable.

11 + 7x = 67 becomes:

7x = 56

We've subtracted 11 from both sides.

We can then isolate x again. By dividing both sides by 7, we get:

x = 8.

Therefore, x = 8.

(a) 36° is equal to (1/5)π radians.

(b) 15 radians is approximately equal to 859.46°.

(c) The angle coterminal to 25/3 that is between 0 and 27 is approximately 14.616.

(a) To convert 36° to radians, we use the conversion factor that 180° is equal to π radians.

36° = (36/180)π = (1/5)π

(b) To convert 15 radians to degrees, we use the conversion factor that π radians is equal to 180°.

15 radians = 15 * (180/π) = 15 * (180/3.14159) ≈ 859.46°

(c) We must add or remove multiples of 2 to 25/3 in order to get an angle coterminal to 25/3 that is between 0 and 27, then we multiply or divide that angle by the necessary range of angles.

25/3 ≈ 8.333

We can add or subtract 2π to get the coterminal angles:

8.333 + 2π ≈ 8.333 + 6.283 ≈ 14.616

8.333 - 2π ≈ 8.333 - 6.283 ≈ 2.050

The angle coterminal to 25/3 that is between 0 and 27 is approximately Between 0 and 27, the angle coterminal to 25/3 is roughly 14.616 degrees.

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a.The number of permutations is:18 × 17 × 16 × ... × 3 × 2 × 1 = 18!

b. The number of permutations is:11! / (5! × 2! × 2!) = 83160.

a. 7 red flags and 11 blue flagsThere are 18 flags in total.

We can choose the first flag in 18 ways, the second flag in 17 ways, the third flag in 16 ways, and so on.

Therefore, the number of permutations is:18 × 17 × 16 × ... × 3 × 2 × 1 = 18!

b. letters of the word ABRACADABRAWe have 11 letters in total.

However, the letter "A" appears 5 times, "B" appears twice, "R" appears twice, and "C" appears once.

Therefore, the number of permutations is:11! / (5! × 2! × 2!) = 83160.

Explanation:We have 6 letters in total.

The word "CARPET" has 2 "E"s, 1 "A", 1 "R", 1 "P", and 1 "T".

Therefore, the number of permutations for the word "CARPET" is:6! / (2! × 1! × 1! × 1! × 1! × 1!) = 180.

The word "CAREER" has 2 "E"s, 2 "R"s, 1 "A", and 1 "C".

Therefore, the number of permutations for the word "CAREER" is:6! / (2! × 2! × 1! × 1! × 1!) = 180.

There are four times as many permutations of the word CARPET as compared to the word CAREER because the word CARPET has only 1 letter repeated twice whereas the word CAREER has 2 letters repeated twice in it.

In general, the number of permutations of a word with n letters, where the letters are not all distinct, is:n! / (p1! × p2! × ... × pk!),where p1, p2, ..., pk are the number of times each letter appears in the word.

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