An internet analytics company measured the number of people watching a video posted on a social media platform. The company found 125 people had watched the video and that the number of people who had watched it was increasing by 30% every 3 hours.
a.) Write an exponential function for the number of people A who had watched the video n hours after the initial observation.
A = _____
b.) A video is said to go "viral" if the number of people who have watched the video exceeds 5 million within 5 days (120 hours). Would this video be considered to have gone viral?
yes or no

Since none of the dot products are equal to zero, none of the angles between the sides of the triangle are 90 degrees. Therefore, the triangle ABC is not a right-angled triangle.

To determine if the triangle ABC is a right-angled triangle, we can check if any of the angles formed by the sides of the triangle are 90 degrees. We can calculate the vectors AB, AC, and BC using the coordinates of the vertices.

Vector AB = B - A = (6-4, 4-(-7), 4-9) = (2, 11, -5)

Vector AC = C - A = (7-4, 10-(-7), -6-9) = (3, 17, -15)

Vector BC = C - B = (7-6, 10-4, -6-4) = (1, 6, -10)

Now, we can calculate the dot products of the vectors to determine the angles between them.

Dot product AB·AC = (2)(3) + (11)(17) + (-5)(-15) = 6 + 187 + 75 = 268

Dot product AB·BC = (2)(1) + (11)(6) + (-5)(-10) = 2 + 66 + 50 = 118

Dot product AC·BC = (3)(1) + (17)(6) + (-15)(-10) = 3 + 102 + 150 = 255

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The given complex number is [tex]z = 343 cis (θ)[/tex]

Find the cube roots of z in terms of θ:Squaring z, we have [tex]z^2 = (343cis(θ))^2= 343^2 cis(2θ)= 117649 cis(2θ)[/tex]

Now, cube root of z is equal to:[tex]∛z = ∛343cis(θ)∛z = ∛343cis(θ + 2πk)[/tex]

Where, k = 0, 1, 2Note: We have used De Moivre's Theorem here.

So,[tex]∛z = 7 cis(θ/3), 7 cis((θ + 2π)/3), 7 cis((θ + 4π)/3)[/tex]Let us plot these roots on the Argand diagram below:Image shows the argand diagram Solution In conclusion,

we have found the cube roots of the given complex number and represented them on a labeled Argand diagram.

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Quadrant I is the first quadrant and the fourth quadrant.

a. Negative x-coordinate and positive y-coordinateIn which quadrant a point has negative x-coordinate and positive y-coordinate?The point which has negative x-coordinate and positive y-coordinate is found in Quadrant II. In this quadrant, x-coordinate is negative and y-coordinate is positive. Hence, the point is located in upper left of the Cartesian plane. The angle made with the positive x-axis by the point is between π/2 and π (90° and 180°).

b. Positive x-coordinate and positive y-coordinateIn which quadrant a point has positive x-coordinate and positive y-coordinate?The point which has positive x-coordinate and positive y-coordinate is found in Quadrant I. In this quadrant, both x-coordinate and y-coordinate are positive. Hence, the point is located in upper right of the Cartesian plane. The angle made with the positive x-axis by the point is between 0° and π/2 (0 and 90°).

c. Positive x-coordinateIn which quadrant a point has positive x-coordinate?The points which has positive x-coordinate are found in Quadrant I and Quadrant IV. In quadrant I, x-coordinate and y-coordinate are both positive while in quadrant IV, x-coordinate is positive and y-coordinate is negative.

Hence, quadrant I is the first quadrant and the fourth quadrant.

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For the given line segments, the vector V can be found by subtracting the coordinates of the initial point A from the coordinates of the final point B. The magnitude of a vector can be calculated using the Pythagorean theorem, which involves finding the square root of the sum of the squares of its components.

To find the vector V given two points A and B, you can subtract the coordinates of point A from the coordinates of point B. Here are the solutions to the two given problems:

1.A=(-5,3) and B=(6,2):

To find vector V, we subtract the coordinates of A from the coordinates of B:

V = (6, 2) - (-5, 3)

= (6 - (-5), 2 - 3)

= (11, -1)

2.A=(2,-8,-3) and B=(-9,4,4):

To find vector V, we subtract the coordinates of A from the coordinates of B:

V = (-9, 4, 4) - (2, -8, -3)

= (-9 - 2, 4 - (-8), 4 - (-3))

= (-11, 12, 7)

Now, to find the magnitude of a vector, you can use the formula:

1.Magnitude of V = [tex]\sqrt(Vx^2 + Vy^2 + Vz^2)[/tex]for a 3D vector.

Magnitude of V = [tex]\sqrt(Vx^2 + Vy^2)[/tex]for a 2D vector.

Let's calculate the magnitudes:

Magnitude of V = [tex]\sqrt(Vx^2 + Vy^2)[/tex] for V = (11, -1)

Magnitude of V = [tex]\sqrt(11^2 + (-1)^2)[/tex]

Magnitude of V = [tex]\sqrt(121 + 1)[/tex]

Magnitude of V = [tex]\sqrt(122)[/tex]

Magnitude of V ≈ 11.045

2.Magnitude of V = [tex]\sqrt(Vx^2 + Vy^2 + Vz^2)[/tex] for V = (-11, 12, 7)

Magnitude of V = [tex]\sqrt((-11)^2 + 12^2 + 7^2)[/tex]

Magnitude of V = [tex]\sqrt(121 + 144 + 49)[/tex]

Magnitude of V =[tex]\sqrt(314)[/tex]

Magnitude of V ≈ 17.720

Therefore, the magnitudes of the vectors are approximately:

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At the end of two years, there will be approximately 26,091 organisms.obtained by applying a growth rate of 12.3% per year to an initial population of 21,900 organisms.

To calculate the population growth after two years, we need to apply the growth rate of 12.3% to the initial population. The growth rate of 12.3% can be expressed as a decimal by dividing it by 100, which gives 0.123.
To find the population at the end of the first year, we multiply the initial population by 1 + growth rate:
P_1 = 21,900 \times (1 + 0.123) = 21,900 \times 1.123 = 24,589.7

P 1 =21,900×(1+0.123)=21,900×1.123=24,589.7.
After rounding to the nearest whole number, the population at the end of the first year is approximately 24,590 organisms.
To find the population at the end of the second year, we repeat the process by multiplying the population at the end of the first year by 1 + growth rate:
P_2 = 24,590 \times (1 + 0.123) = 24,590 \times 1.123 = 26,090.37

P 2=24,590×(1+0.123)=24,590×1.123=26,090.37.
Again, after rounding to the nearest whole number, the population at the end of the second year is approximately 26,091 organisms.
Therefore, at the end of two years, there will be approximately 26,091 organisms.

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Statistical testing examines the relationship between the independent variable (IV) and the dependent variable (DV) in order to determine if there is a significant association

Statistical testing examines the relationship between variables to determine if there is a significant association. In this scenario, the IV is the intervention type, which can be either writing focused or no intervention. The DV is the improved overall writing, categorized as either success (writing improved) or failure (no improvement).

To perform a chi-square test, we need to state a hypothesis. A one-tailed hypothesis suggests the direction of the expected relationship. Let's assume we expect the writing focused intervention to lead to improved overall writing (success). Our one-tailed hypothesis would be:

H₀: The intervention type has no effect on the improvement of overall writing (success and failure are equally likely).

H₁: The writing focused intervention leads to improved overall writing (success is more likely than failure).

To calculate the chi-square statistic, we need the observed frequencies for the different combinations of IV and DV. The observed frequencies given are 40, 10, 60, and 90. However, it's not clear how these frequencies are distributed across the different categories. Without the specific distribution, it is not possible to calculate the chi-square statistic.

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The function G(n) in terms of f(n) is G(n) = 2/π² (f(n) - f²(n)).

To find the function G(n) in terms of f(n) based on the given expression, we substitute f(n) into the formula for G(n):

G(n) = 2/π² (Sin nπ/2 - Sin² nπ/2)

Replacing Sin nπ/2 with f(n), we have:

G(n) = 2/π² (f(n) - Sin² nπ/2)

Since f(n) is defined as f(n) = Sin nπ/2, we can simplify further:

G(n) = 2/π² (Sin nπ/2 - Sin² nπ/2)

Now we can substitute f(n) = Sin nπ/2 into the equation:

G(n) = 2/π² (f(n) - f²(n))

Therefore, the function G(n) in terms of f(n) is G(n) = 2/π² (f(n) - f²(n)).

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Given that the Hydrogen ion Concentration of RC Cola is 4.7863 × 10⁻³.

We need to find its pH.

As we know that,

pH = -log[H⁺]

Hence, pH = -log[4.7863 × 10⁻³]pH = 2.320The pH of RC Cola is 2.32.

Therefore, option B is correct.

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The statement asserts that there is at least one student who listens to all of their professors.

The statement "Some students listen to every one of their professors" can be understood as follows:

1. Sx: x is a student.

This predicate defines Sx as the property of x being a student. It indicates that x belongs to the group of students.

2. Pxy: x is a professor of y.

This predicate defines Pxy as the property of x being a professor of y. It indicates that x is the professor of y.

3. Lxy: x listens to y.

This predicate defines Lxy as the property of x listening to y. It indicates that x pays attention to or follows the teachings of y.

The statement states that there exist some students who listen to every one of their professors. This means that there is at least one student who listens to all the professors they have.

The logical representation of this statement would be:

∃x(Sx ∧ ∀y(Pyx → Lxy))

Breaking down the logical representation:

∃x: There exists at least one x.

(Sx: x is a student): This x is a student.

∀y(Pyx → Lxy): For every y, if y is a professor of x, then x listens to y.

In simpler terms, the statement asserts that there is at least one student who listens to all of their professors.

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The stream function, denoted by ψ, can be determined from the velocity potential, denoted by ф, by taking the partial derivatives with respect to the coordinates. In this case, the velocity potential is given as ф = [tex]3xy^2 - x^3[/tex]. To find the stream function, we will calculate the partial derivatives and rearrange the equations.

The stream function, denoted by ψ, is related to the velocity potential through the following equations:

ψ_x = -ф_y   and   ψ_y = ф_x

Taking the partial derivative of ф with respect to y, we have:

ф_y = [tex]3x(2y) - 0 = 6xy[/tex]

Equating this to -ψ_x, we get:

-ψ_x = [tex]6xy[/tex]

Integrating this equation with respect to x yields:

ψ = [tex]-3xy^2 + g(y)[/tex]

Here, g(y) represents an arbitrary function of y that arises due to the integration process.

Similarly, taking the partial derivative of ф with respect to x, we have:

ф_x = [tex]3y^2 - 3x^2[/tex]

Equating this to ψ_y, we get:

ψ_y = [tex]3y^2 - 3x^2[/tex]

Integrating this equation with respect to y yields:

ψ = [tex]y^3 - 3xy^2 + f(x)[/tex]

Here, f(x) represents an arbitrary function of x.

Combining the two expressions for ψ, we have:

ψ = [tex]-3xy^2 + g(y) = y^3 - 3xy^2 + f(x)[/tex]

Since g(y) and f(x) are arbitrary functions, we can set them to zero.

Therefore, the stream function for the given flow field is:

ψ = [tex]y^3 - 3xy^2[/tex]

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The histogram displays the distribution of geode masses, with the x-axis representing the mass intervals and the y-axis representing the frequency of geodes within each interval.

To create a histogram of the mass of geodes found at a volcanic site, follow these steps:

Determine the range of the data. The minimum value is 0.8 kg, and the maximum value is 26.8 kg.

Decide on the number of bins or intervals for the histogram. Let's choose 8 bins for this example.

Calculate the bin width by dividing the range by the number of bins. In this case, the bin width is (26.8 - 0.8) / 8 = 3.375 kg.

Create the intervals for the bins by starting from the minimum value and incrementing by the bin width. The intervals are:

0.8 - 4.175 kg

4.175 - 7.95 kg

7.95 - 11.725 kg

11.725 - 15.5 kg

15.5 - 19.275 kg

19.275 - 23.05 kg

23.05 - 26.825 kg

Count the number of geodes that fall within each interval. From the given data, you can determine the frequencies for each interval.

Create the histogram by representing the intervals on the x-axis and the frequencies on the y-axis. Use bars of different lengths to represent the frequencies. Label the axes and provide a title for the histogram.

Here is  the histogram-

         Frequency

          |  

  7       |   *    

  6       |  

  5       |  

  4       |  

  3       |   *

  2       |   **

  1       |   *

  0       |____________________

          0.8   7.95   15.5   23.05   26.825  (kg)

The histogram displays the distribution of geode masses, with the x-axis representing the mass intervals and the y-axis representing the frequency of geodes within each interval. The bars depict the frequencies for each interval, showing the pattern of the mass distribution.

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The answer is , it can be concluded that if [tex]\(\int_a^bf(x)dx=0\)[/tex]and (f(x)) is continuous, then (a=b) is a statement that is True.

The statement, "If[tex]\(\int_a^bf(x)dx=0\)[/tex] and [tex]\(f(x)\)[/tex] is continuous, then (a=b) is a statement that is True.

If[tex]\(\int_a^bf(x)dx=0\)[/tex]and (f(x)) is continuous, then this means that the area under the curve is equal to 0.

The reason that the integral is equal to zero can be seen graphically, since the areas above and below the (x)-axis must cancel out to result in an integral of 0.

Since (f(x)) is a continuous function, it doesn't have any jump discontinuities on the interval ([a,b]),

which means that it is either always positive, always negative, or 0.

This rules out the possibility that there are two areas of opposite sign that can cancel out in order to make the integral equal to zero.

Thus, if the area under the curve is equal to zero, then the curve must lie entirely on the (x)-axis,

which means that the only way for this to happen is if \(a=b\).

Hence, it can be concluded that if [tex]\(\int_a^bf(x)dx=0\)[/tex]and (f(x)) is continuous, then (a=b) is a statement that is True.

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The number of solutions to the equation x1 + x2 + x3 + x4 + x5 = 15 in positive integers x1, x2, x3, x4, and x5, not exceeding 6, is 4

To determine the number of solutions of the equation x1 + x2 + x3 + x4 + x5 = 15 in positive integers x1, x2, x3, x4, and x5, not exceeding 6, we can use the concept of generating functions.

We can represent each variable (x1, x2, x3, x4, and x5) as a polynomial in the generating function. Since the values cannot exceed 6, the polynomial for each variable can be expressed as:

x1: 1 + x + x^2 + x^3 + x^4 + x^5 + x^6

x2: 1 + x + x^2 + x^3 + x^4 + x^5 + x^6

x3: 1 + x + x^2 + x^3 + x^4 + x^5 + x^6

x4: 1 + x + x^2 + x^3 + x^4 + x^5 + x^6

x5: 1 + x + x^2 + x^3 + x^4 + x^5 + x^6

To find the number of solutions, we need to find the coefficient of x^15 in the product of these polynomials.

Multiplying the polynomials:

(x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)^5

Expanding this expression and finding the coefficient of x^15, we get:

Coeff(x^15) = 5 + 10 + 10 + 10 + 5 + 1 = 41

Therefore, the number of solutions to the equation x1 + x2 + x3 + x4 + x5 = 15 in positive integers x1, x2, x3, x4, and x5, not exceeding 6, is 41.

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the dimensions of the rectangular playground with the largest area would be a square with each side measuring approximately 33.33 meters.

To find the dimensions of the rectangular playground with the largest area using 100 meters of fencing, we can apply the concept of optimization. The maximum area of a rectangle can be obtained when it is a square. Therefore, we can aim for a square playground.

Considering a square playground, let's denote the length of each side as "s." Since we have three sides of fencing, two sides will be parallel and equal in length, while the third side will be perpendicular to them. Hence, the perimeter of the playground can be expressed as P = 2s + s = 3s.

Given that we have 100 meters of fencing, we can set up the equation 3s = 100 to find the length of each side. Solving for s, we get s = 100/3.

Thus, the dimensions of the rectangular playground with the largest area would be a square with each side measuring approximately 33.33 meters.

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The equation that is required to be solved is: [tex]$$\sqrt{2t} 2t - 1 + t = 53.56$$$$\sqrt{3t}+ 3 = 5x$$[/tex]

Solving the first equation: [tex]$$\begin{aligned}\sqrt{2t} 2t - 1 + t &= 53.56\\2t^2 + t - 53.56 &= 1\\2t^2 + t - 54.56 &= 0\end{aligned}$$[/tex]

Now we can apply the quadratic formula to solve for t. The quadratic formula is:[tex]$$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$[/tex]

Using the quadratic formula for the equation above, we can substitute the values of a, b and c as follows: [tex]$$\begin{aligned}a &= 2\\b &= 1\\c &= -54.56\\\end{aligned}$$[/tex]

Substituting the values into the quadratic formula gives us:[tex]$$t=\frac{-1 \pm \sqrt{1-4(2)(-54.56)}}{2(2)}$$$$t=\frac{-1 \pm \sqrt{1+436.48}}{4}$$$$t=\frac{-1 \pm \sqrt{437.48}}{4}$$[/tex]

The solutions are:[tex]$$t_1 = \frac{-1 + \sqrt{437.48}}{4}$$$$t_2 = \frac{-1 - \sqrt{437.48}}{4}$$[/tex]

Calculating t1 and t2 using a calculator gives:[tex]$$t_1 \approx 3.743$$$$t_2 \approx -7.344$$[/tex]

However, since we are dealing with time, a negative value for t is not acceptable. Therefore, the only solution is

[tex]$$t = t_1$$[/tex]

Substituting t into the second equation gives: [tex]$$\sqrt{3(3.743)}+ 3 = 5x$$$$\sqrt{11.229}+ 3 = 5x$$$$5x = \sqrt{11.229}+ 3$$$$5x = 6.345$$$$x \approx 1.269$$[/tex]

Therefore, the solution to the equations is[tex]$$t \approx 3.743$$and$$x \approx 1.269$$[/tex]

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The half-angle formulas are used to determine the exact values of sine, cosine, and tangent of an angle. These formulas are generally used to simplify trigonometric equations involving these three functions.

The half-angle formulas are as follows:

[tex]sin(θ/2) = ±sqrt((1 - cos(θ))/2)cos(θ/2) = ±sqrt((1 + cos(θ))/2)tan(θ/2) = sin(θ)/(1 + cos(θ)) = 1 - cos(θ)/sin(θ)[/tex]

To determine the exact values of the sine, cosine, and tangent of 15⁰, we can use the half-angle formula for sin(θ/2) as follows: First, we need to convert 15⁰ into 30⁰ - 15⁰ using the angle subtraction formula, i.e.

[tex],sin(15⁰) = sin(30⁰ - 15⁰[/tex]

Next, we can use the half-angle formula for sin(θ/2) as follows

:sin(θ/2) = ±sqrt((1 - cos(θ))/2)Since we know that sin(30⁰) = 1/2 and cos(30⁰) = √3/2,

we can write:

[tex]sin(15⁰) = sin(30⁰ - 15⁰) = sin(30⁰)cos(15⁰) - cos(30⁰)sin(15⁰)= (1/2)(√6 - 1/2) - (√3/2)(sin[/tex]

Multiplying through by 2 and adding sin(15⁰) to both sides gives:

2sin(15⁰) + √3sin(15⁰) = √6 - 1

The exact values of sine, cosine, and tangent of 15⁰ using the half-angle formulas are:

[tex]sin(150) = (√6 - 1)/(2 + √3)cos(150) = -√18 + √6 + 2√3 - 2tan(15⁰) = (-1/2)(2 + √3)[/tex]

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The area and distance are as follows::

(a) The area of ABEF can be found by using the formula for the area of a parallelogram: Area = base × height. Since ABEF shares a base with ABCD and has the same height as the distance between the parallel lines, the area of ABEF is equal to the area of ABCD, which is 5 square units.

(b) Similarly, the area of ABGH can also be determined as 8 square units using the same approach as in part (a). Both ABEF and ABGH share a base with ABCD and have the same height as the distance between the parallel lines.

(c) Given that AB = 2 units, we can find the distance between the parallel lines by using the formula for the area of a parallelogram:

Area = base × height

Since the area of ABCD is 5 square units and the base AB is 2 units, the height is:

height = Area / base = 5 / 2 = 2.5 units

Therefore, the distance between the parallel lines is 2.5 units.

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The linear approximation of f(x) about x = 2 is L(x) = 8 + 24(x - 2). Using this, f(3) is approximately equal to 32. The second-order approximation of f(x) about x = 2 is Q(x) = 8 + 24(x - 2) + (1/2)(44)[tex](x - 2)^2[/tex]. Using this, f(3) is approximately equal to 54.

To obtain the linear approximation of the function f(x) = x^4 - 2x^2 about the point x = 2, we can use the concept of a tangent line. The linear approximation is given by:

L(x) = f(a) + f'(a)(x - a),

where a is the point of approximation, f(a) is the value of the function at a, and f'(a) is the derivative of the function evaluated at a.

Linear Approximation:

Let's calculate the linear approximation of f(x) about x = 2.

a = 2,

f(a) = f(2)

[tex]= (2^4) - 2(2^2)[/tex]

= 16 - 8

= 8,

[tex]f'(x) = 4x^3 - 4x[/tex], (derivative of f(x)),

f'(a) = f'(2)

[tex]= 4(2^3) - 4(2)[/tex]

= 32 - 8

= 24.

Using these values, we have:

L(x) = 8 + 24(x - 2).

Computing f(3) using the linear approximation:

To compute f(3) using the linear approximation, substitute x = 3 into L(x):

L(3) = 8 + 24(3 - 2)

= 8 + 24

= 32.

Second-Order Approximation:

The second-order approximation takes into account the first and second derivatives of the function. It is given by:

[tex]Q(x) = f(a) + f'(a)(x - a) + (1/2)f''(a)(x - a)^2,[/tex]

where f''(a) is the second derivative of the function evaluated at a.

To compute the second-order approximation of f(x) about x = 2:

a = 2,

f(a) = f(2)

= 8,

f'(a) = f'(2)

= 24,

[tex]f''(x) = 12x^2 - 4,[/tex] (second derivative of f(x)),

f''(a) = f''(2)

[tex]= 12(2^2) - 4[/tex]

= 48 - 4

= 44.

Using these values, we have:

[tex]Q(x) = 8 + 24(x - 2) + (1/2)(44)(x - 2)^2.[/tex]

Computing f(3) using the second-order approximation:

To compute f(3) using the second-order approximation, substitute x = 3 into Q(x):

[tex]Q(3) = 8 + 24(3 - 2) + (1/2)(44)(3 - 2)^2[/tex]

= 8 + 24 + 22

= 54

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We are given an arithmetic sequence with the common ratio [tex]\(r = 25\)[/tex] and the sum of the first seven terms [tex]\(S_7 = 70\)[/tex]. We are asked to find the first term [tex]\(a_1\)[/tex] and the common difference [tex]\(d\)[/tex] of the sequence.

In an arithmetic sequence, each term can be represented as [tex]\(a_n = a_1 + (n-1)d\)[/tex], where [tex]\(a_n\)[/tex] is the [tex]\(n\)th[/tex] term, [tex]\(a_1\)[/tex] is the first term, [tex]\(d\)[/tex] is the common difference, and [tex]\(n\)[/tex] is the position of the term.

From the given information, we have [tex]\(r = 25\)[/tex] and [tex]\(S_7 = 70\)[/tex]. The sum of the first seven terms is given by the formula [tex]\(S_7 = \frac{n}{2}(a_1 + a_7)\)[/tex].

Substituting the values into the formula, we get:

[tex]\(70 = \frac{7}{2}(a_1 + a_1 + 6d)\)\(70 = \frac{7}{2}(2a_1 + 6d)\)\\\(70 = 7(a_1 + 3d)\)\\\(10 = a_1 + 3d\[/tex] (Dividing both sides by 7)

Since [tex]\(r = 25\) and \(a_1 = d\)[/tex], we can substitute these values into the equation:

[tex]\(10 = a_1 + 3a_1\)\\\(10 = 4a_1\)\\\(a_1 = \frac{10}{4} = 2.5\)[/tex]

Therefore, the first term [tex]\(a_1\)[/tex] of the arithmetic sequence is[tex]\(2.5\)[/tex]and the common difference [tex]\(d\)[/tex] is also [tex]\(2.5\)[/tex].

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1. The probability that the first chip drawn is white and the second chip drawn is blue is 1/36.

2. The probability that the first two chips drawn are both red is 1/6.

3. The probability that the first four chips drawn are all yellow is 1/1512.

To calculate the probabilities, we need to count the number of favorable outcomes and the total number of possible outcomes.

1. Probability of drawing a white chip followed by a blue chip:

The total number of chips is 9. Among them, there is 1 white chip and 2 blue chips. The probability of drawing a white chip first is 1/9. After drawing a white chip, there are 8 chips remaining, including 2 blue chips. So, the probability of drawing a blue chip second, without replacement, is 2/8. To find the probability of both events occurring, we multiply the individual probabilities:

P(white and blue) = (1/9) * (2/8) = 1/36

2. Probability of drawing two red chips:

The total number of chips is 9. Among them, there are 4 red chips. The probability of drawing a red chip first is 4/9. After drawing a red chip, there are 8 chips remaining, including 3 red chips. So, the probability of drawing a second red chip, without replacement, is 3/8. To find the probability of both events occurring, we multiply the individual probabilities:

P(two red) = (4/9) * (3/8) = 1/6

3. Probability of drawing four yellow chips:

The total number of chips is 9. Among them, there are 2 yellow chips. The probability of drawing a yellow chip first is 2/9. After drawing a yellow chip, there are 8 chips remaining, including 1 yellow chip. So, the probability of drawing a second yellow chip, without replacement, is 1/8. Similarly, the probabilities of drawing the third and fourth yellow chips, without replacement, are 1/7 and 1/6, respectively. To find the probability of all four events occurring, we multiply the individual probabilities:

P(four yellow) = (2/9) * (1/8) * (1/7) * (1/6) = 1/1512

Therefore:

1. The probability that the first chip drawn is white and the second chip drawn is blue is 1/36.

2. The probability that the first two chips drawn are both red is 1/6.

3. The probability that the first four chips drawn are all yellow is 1/1512.

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The simplified expression is:

1 + csc(0)

0

Which is undefined.

Starting with the given expression:

1 - csc(0)cos(0)

1 + csc(0)(1 - csc(0))

We can recall the following trigonometric identities:

csc(0) = 1/sin(0) = undefined

cos(0) = 1

Since csc(0) is undefined, we cannot directly substitute it into the expression. However, we can use the fact that sin(0) = 0 to simplify the expression.

1 - (undefined)(1)

1 + (undefined)(1 - undefined)

Since the denominator contains an undefined term, we need to find a way to remove it. To do this, we can multiply both the numerator and denominator by the conjugate of the denominator, which is (1 + csc(0)).

(1 - undefined)(1 + csc(0))(1)

(1 + undefined)(1 - csc(0))(1 + csc(0))

Simplifying the numerator gives us:

(1 - undefined)(1 + csc(0)) = 1 + csc(0)

And simplifying the denominator gives us:

(1 + undefined)(1 - csc(0))(1 + csc(0)) = (1 - csc^2(0))(1 + csc(0)) = -sin^2(0)(1 + csc(0))

Substituting sin(0) = 0, we get:

-0(1 + csc(0)) = 0

Therefore, the simplified expression is:

1 + csc(0)

0

Which is undefined.

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The equation is given by: y = -1 / b + cos(x)Here, 0 ≤ x ≤ 2π and 2 ≤ b ≤ 6.The question asks to find the lowest point of the graph. The value of b determines the vertical displacement of the graph.

As the value of b increases, the graph shifts downwards. Thus, as b increases, the lowest point of the graph also moves down. The graph can be plotted for different values of b. The graph can be analyzed to find the point where it reaches its minimum value.

For b = 2, the graph is as shown below: For b = 6, the graph is as shown below:

The graphs clearly show that as the value of b increases, the graph shifts downwards. This is consistent with the equation as the vertical displacement is controlled by the value of b.

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The lengths of XY and YZ of the triangle are:

XY = 6 cm

YZ = 8 cm

We have that:

The ratio of the area of ΔWXY to the area of ΔWZY is 3:4.

The area of ΔWXZ is 112 cm² and WY = 16 cm.

Thus,

Total of the ratio = 3 + 4 = 7

area of ΔWXY = 3/7 * 112 = 48 cm²

area of ΔWZY = 4/7 * 112 = 64 cm²

Area of triangle = 1/2 * base * height

For ΔWXY:

area of ΔWXY = 1/2 * XY * WY

48 = 1/2 * XY * 16

48 = 8XY

XY = 48/8

XY = 6 cm

For ΔWZY:

area of ΔWZY = 1/2 * YZ * WY

64 = 1/2 * YZ * 16

64 = 8YZ

YZ = 64/8

YZ = 8 cm

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The correct statement is:

c. If A is a square matrix, then A is invertible if A³ = I, then A⁻¹ = A².

a. If the homogeneous system AX = 0 has a non-zero solution, then the columns of matrix A are linearly dependent.

This statement is true. If the homogeneous system AX = 0 has a non-zero solution, it means there exists a non-zero vector X such that AX = 0. In other words, the columns of matrix A can be combined linearly to produce the zero vector, indicating linear dependence.

b. If the homogeneous system AX = 0 has a non-zero solution, then the columns of matrix A are linearly independent.

This statement is false. The correct statement is the opposite: if the homogeneous system AX = 0 has a non-zero solution, then the columns of matrix A are linearly dependent (as mentioned in statement a).

c. If A is a square matrix, then A is invertible if A³ = I, then A⁻¹ = A².

This statement is false. The correct statement should be: If A is a square matrix and A³ = I, then A is invertible and A⁻¹ = A². If a square matrix A raised to the power of 3 equals the identity matrix I, it implies that A is invertible, and its inverse is equal to its square (A⁻¹ = A²).

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When the value of the variable = 2 the value of  W = 3.When the value of one quantity increases with respect to decrease in other or vice-versa, then they are said to be inversely proportional. It means that the two quantities behave opposite in nature. For example, speed and time are in inverse proportion with each other. As you increase the speed, the time is reduced.

In the problem it's given that "y varies inversely as x," and "when x = 6, then y = 4."

We need to find y when x = 7, we can use the formula for inverse variation:

y = k/x  where k is the constant of variation.

To find the value of k, we can plug in the given values of x and y:

4 = k/6

Solving for k:

k = 24

Now, we can plug in k and the value of x = 7 to find y:

y = 24/7

Answer: y = 24/7

Function for the inverse variation between W and square of 2 can be written as follows,

W = k/(2)^2 = k/4

It is given that when 12 = 3, W = 3,

So k/4 = 3

k = 12

Now, we need to find W when variable = 2,

Thus,

W = k/4

W = 12/4

W = 3

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The value of the speed boat after 7 years is found to be  $26,179.

The value of the boat after 7 years can be calculated by using the formula for continuous compound interest:

[tex]A = Pe^(rt)[/tex]

where A is the final amount,

P is the initial amount,

e is Euler's number (approximately 2.71828),

r is the annual interest rate as a decimal, and

t is the time in years.

Here, the initial amount is $13,000, the annual interest rate is 10%, and the time is 7 years.

So, we have:

[tex]A = 13000e^(0.1*7)\\A = 13000e^(0.7)[/tex]

A = $26179.23

Therefore, the value of the speed boat after 7 years is $26,179. Rounded to the nearest dollar, amount is is $26179.

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As time (t) increases, the exponential term decreases, causing the value (V) of the Escalade to decrease.

the initial value of the Escalade is $85,000.

a) The value of an Escalade decreases over time. This can be observed from the exponential function V(t) = 75(0.798)^t+10. The term (0.798)^t represents exponential decay since the base (0.798) is between 0 and 1. As time (t) increases, the exponential term decreases, causing the value (V) of the Escalade to decrease.

b) The initial value of the Escalade can be found by plugging in t = 0 into the equation V(t) = 75(0.798)^t+10:

V(0) = 75(0.798)^0+10

V(0) = 75(1) + 10

V(0) = 75 + 10

V(0) = 85

Therefore, the initial value of the Escalade is $85,000.

c) To determine the value of the Escalade in 5 years, we need to substitute t = 5 into the equation V(t):

V(5) = 75(0.798)^5+10

V(5) ≈ 75(0.512) + 10

V(5) ≈ 38.4 + 10

V(5) ≈ 48.4

Therefore, the value of the Escalade after 5 years is approximately $48,400.

d) The equation of the asymptote is y = 10. In this situation, the asymptote represents the minimum value that the Escalade's value will approach over time. As t approaches infinity, the exponential term (0.798)^t approaches 0, and the value of the Escalade gets closer and closer to the asymptote value of $10,000. This suggests that the Escalade's value will never reach or go below $10,000.

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a) The inverse of function is,

⇒ f⁻¹(x) = ±√(x - 2).

b) The domain of f(x) is all real numbers,

c) Graph f, f', and y=x on the same coordinate axes are shown in image.

To find the inverse of f(x) = x² + 2, we can start by rewriting it as,

y = x² + 2.

Then, we can switch the roles of x and y, and solve for y:

x = y² + 2

x - 2 = y²

y = ±√(x - 2)

So the inverse of f(x) is,

⇒ f⁻¹(x) = ±√(x - 2).

So f(f⁻¹(x)) = x, which means that f⁻¹(x) is indeed the inverse of f(x).

Moving on to part (b), the domain of f(x) is all real numbers,

Since, x² + 2 is defined for any real value of x.

The range of f(x) is all real numbers greater than or equal to 2, since x² is always non-negative and adding 2 to it makes it at least 2.

As for the domain and range of f⁻¹(x), the domain is all real numbers greater than or equal to 2 (since √(x - 2) can only be real when x - 2 is non-negative), and the range is all real numbers.

Finally, for part (c), let's graph f(x), f'(x) , and y = x on the same coordinate axes.

As you can see, f(x) is a parabola opening upward, f'(x) is a straight line (2x), and y = x is a diagonal line.

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The correct option the third one, the value of x is x = -9,

We can see that we have an isosceles triangle. Then two of the interior angles have the measure ∠2, and the other angle has the measure of 60°.

We know that the sum of the interior angles is equal to 180°, then we can write:

60° + 2*∠2 = 180°

60°  + 2*(x + 69) = 180°

2*(x + 69) = 180 - 60 = 120

x + 69 = 120/2

x = 60 - 69

x = -9

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The answers to the given problem are: (a) 24.32 years (b) 28.47 years.

Given data: Amount invested = $2500

Interest rate = r = 0.0285 (compounded continuously)

Let's calculate the time required for the amount to double and triple.

Doubling Time: The formula for doubling time (t2) is given by;

ln(2) / r = t2

Where, ln(2) = 0.693

So,t2 = 0.693 / 0.0285t2 = 24.32 years

Thus, the time required for the amount to double is 24.32 years.

Tripling Time: The formula for tripling time (t3) is given by;

ln(3) / r = t3So,t3 = ln(3) / r = 0.8109 / 0.0285t3 = 28.47 years

Thus, the time required for the amount to triple is 28.47 years.

Therefore, the answers to the given problem are: (a) 24.32 years (b) 28.47 years.

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