Determine whether the infinite series m=1 Σ 4+3^m/5^mconverges or diverges, and if it converges, it find sum:1. converges with sum = 11/4 2. series diverges 3. converges with sum = 23/84. converges with sum = 21/8 5. converges with sum = 19/8

Answer: 32 m

Step-by-Step Explanation:

To solve this problem, we can use a combination formula. We want to choose 10 coins from a total of 80 nickels and 100 pennies, First, we need to determine how many ways we can choose 0-10 nickels. We can represent this with the following formula:

(number of ways to choose 0 nickels) + (number of ways to choose 1 nickel) + ... + (number of ways to choose 10 nickels) To find the number of ways to choose a certain number of nickels, we can use combinations. For example, the number of ways to choose 3 nickels from 80 is:

80C3 = (80!)/(3!(80-3)!) = 82,160

Using this method, we can find the number of ways to choose 0-10 nickels:

(number of ways to choose 0 nickels) = 1

(number of ways to choose 1 nickel) = 80C1 = 80

(number of ways to choose 2 nickels) = 80C2 = 3,160

(number of ways to choose 3 nickels) = 80C3 = 82,160

(number of ways to choose 4 nickels) = 80C4 = 1,484,480

(number of ways to choose 5 nickels) = 80C5 = 17,259,280

(number of ways to choose 6 nickels) = 80C6 = 119,759,850

(number of ways to choose 7 nickels) = 80C7 = 524,512,800

(number of ways to choose 8 nickels) = 80C8 = 1,719,596,080

(number of ways to choose 9 nickels) = 80C9 = 41,079,110

(number of ways to choose 10 nickels) = 80C10 = 1,028,671


Therefore, there are 13,958,883,175 ways to choose 10 coins from a bank containing 80 identical nickels and 100 identical pennies, To find the number of ways to choose 10 coins from a bank containing 80 identical nickels and 100 identical pennies, we'll use a combination formula. However, since the coins are identical, we can simplify the problem by counting the number of ways to choose a certain amount of nickels and then filling the rest of the 10 coins with pennies.

1. Choose 0 nickels and 10 pennies: This is only one way, since all coins are identical.
2. Choose 1 nickel and 9 pennies: This is also one way.
3. Choose 2 nickels and 8 pennies: This is one way as well.

1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 11 ways

There are 11 different ways to choose 10 coins from a bank containing 80 identical nickels and 100 identical pennies.

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The domain of the which the function is increasing from the graph is

The domain of the which the function is increasing from the graph is determined by observing when the graph is starts to point up wards

Examining the graph points after x = -4 is the starting point.

Since the graph has arrow ends the end point is not seen on the graph in this case we represent it with infinity ∞

These points are not inclusive as we have points after -4 but not -4 itself and points tending to infinity. We represent these points mathematically as

(-4 ∞)

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You can proceed with evaluating the integral, depending on the specific form of the function SA(x).

First, let's rewrite the given information to clarify the problem:

(a) Let R be the region enclosed by the lines y = x, y = 6 - 2x, and y = 53. We want to evaluate the double integral of the function SA(x) over the region R.

To find the limits of integration, we need to determine the intersection points of the given lines. Let's find the intersection of y = x and y = 6 - 2x:

x = 6 - 2x
3x = 6
x = 2
y = 2

The intersection point is (2, 2).

Now, let's evaluate the double integral of SA(x) over the region R. We can set up the integral as follows:

∬_R SA(x) dA = ∫(0 to 2) ∫(x to 6 - 2x) SA(x) dy dx

Now you can proceed with evaluating the integral, depending on the specific form of the function SA(x).

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The surface area of the prism is 684 in².

We have,

Rectangular prism:

Surface area = 2lw + 2lh + 2wh,

where l, w, and h are the lengths of the three sides.

Now,

l = 12

w = 15

h = 6

Substituting.

Surface area

= 2lw + 2lh + 2wh

= 2 x 12 x 15 + 2 x 12 x 6 + 2 x 15 x 6
= 360 + 144 + 180

= 684 in²

Thus,

The surface area of the prism is 684 in².

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The smallest positive integer n, such that 1 / 9 is a terminating decimal and n contains 9 is 4, 096.

Finite digits terminating after the decimal point represent what are known as "terminating decimals". This type of decimal is characterized by their limited representation which comes to an end after a specific number of digits.

The smallest positive integer to satisfy the conditions, of the terminating decimals would be in the form 2 ^ r 5 ^ s.

We can then solve for the smallest positive integer n, to be:

= 2 ¹² x 5 ⁰

= 4 ,096

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A graph that represent the quadratic equation y = -x² + 4x + 21 is shown in the image attached below.

In Mathematics and Geometry, the graph of a quadratic function would always form a parabolic curve because it is a u-shaped. Based on the given quadratic function, we can logically deduce that the graph would be a downward parabola because the coefficient of x² is negative and the value of "a" is lesser than zero (0).

Since the leading coefficient (value of a) in the given quadratic function y = -x² + 4x + 21 is negative 1, we can logically deduce that the parabola would open downward and the solution would be represented by the following x-intercepts (zeros or roots);

Ordered pair = (-3, 0)

Ordered pair = (0, 7)

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Answer:

y = -1/2 x + 1

Step-by-step explanation:

You can find the gradient by finding the rise/run. It is -1/2 as seen in the equation, and then then slope needs to be moved upwards by one to meet the correct y coordinates. Remember y = mx + c.

For part a, the confidence interval for the population proportion of college students who work to pay for tuition and living expenses is 0.42 to 0.52, with a certain level of confidence (usually 95% or 99%).
For part b, the confidence interval is slightly wider and ranges from 0.41 to 0.53. This could be due to a larger sample size or a lower level of confidence.
For part c, as the confidence level increases from 95% to 99%, the margin of error becomes larger.

a. To calculate the confidence interval for the population proportion of college students who work to pay for tuition and living expenses, we use the given range of 0.42 to 0.52. This interval indicates that we can be confident that the true population proportion falls between 42% and 52% of college students. This means that we are 95% confident that the true population proportion falls within this interval based on the sample data.

b. Similarly, for the second provided confidence interval, we use the given range of 0.41 to 0.53. This interval indicates that we can be confident that the true population proportion falls between 41% and 53% of college students.

c. When the confidence level is increased, the margin of error becomes larger. This is because a higher confidence level requires a wider interval to ensure that the true population proportion falls within the specified range with greater certainty. This is because a higher level of confidence requires a wider interval to capture the true population proportion. As a result, the precision of the estimate decreases as the margin of error increases.

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To achieve the least area, create two separate square enclosures, each with a side length of 25 m. For the greatest area, make one enclosure with a side length of 49 m and another with a side length of 1 m.

To determine the plans for the least and greatest areas using 200 m of fencing, we'll consider two cases: one square enclosure and two square enclosures.

Case 1: One square enclosure
Perimeter = 200 m
Since the perimeter of a square is 4 * side length (s), we have:
200 = 4 * s
s = 50 m

Area of one square enclosure = s^2 = 50^2 = 2500 m^2

Case 2: Two square enclosures
Let s1 and s2 be the side lengths of the two square enclosures.
Perimeter = 200 m
4 * (s1 + s2) = 200
s1 + s2 = 50
Since the area of a square is side length squared, we have:
Area = s1^2 + s2^2

To minimize the area, make the side lengths equal:
s1 = s2 = 25 m
Minimum area = 2 * (25^2) = 2 * 625 = 1250 m^2

To maximize the area, make one side length as large as possible while keeping the perimeter constraint:
s1 = 49 m, s2 = 1 m
Maximum area = 49^2 + 1^2 = 2401 + 1 = 2402 m^2

Therefore, to achieve the least area, create two separate square enclosures, each with a side length of 25 m. For the greatest area, make one enclosure with a side length of 49 m and another with a side length of 1 m.

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When polling individuals about who they will likely vote for in the next election, an additional question should be asked about their political affiliation or ideology to avoid a biased sample.

This will ensure that the sample is representative of the entire population, rather than just a particular group or demographic that may have a certain tendency to vote for a particular candidate. By asking about political affiliation or ideology, the pollster can account for any potential biases that may exist within the sample and ensure that the results are more accurate and reliable.

To avoid a biased sample when polling individuals about their likely vote in the next election, an additional question that should be asked is: "Did you vote in the previous election?" This helps to ensure that you are including opinions from both regular voters and those who might not have participated before, providing a more accurate representation of the electorate.

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Answer:

√37 is about 6.08, so N is the correct letter (3 is the correct choice).

The length of AB is √63

The measure of ∠A is 49°

The measure of ∠C is 41°

The measure of ∠B is 90°

We have,

1)

Using the Pythagorean theorem,

Hypotenuse = AC

Base = BC

Height = AB

AC² = BC² + AB²

AC² - BC² = AB²

AB² = 144 - 81

AB² = 63

AB = √63

AB = 7.9

AB = 8

2)

Sin A = BC/AC

Sin A = 9/12

Sin A = 3/4

A = [tex]sin^{-1}0.75[/tex]

A = 48.59

A = 49°

3)

Sin C = AB/AC

Sin C = √63/12

C = [tex]sin^{-1}0.66[/tex]

C = 41°

4)

∠B = 90

Thus,

The length of AB is √63

The measure of ∠A is 49°

The measure of ∠C is 41°

The measure of ∠B is 90°

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In = [ (22 +16) = dx, where n= 1,2,3,... is a positive integer, the expressions for An and Bn are: An = 4 Bn = 36n^2 + 124n + 144

To solve this problem, we will use integration by parts. Let's start by setting u = x^2 + 16 and dv = dx.

Then we have du = 2x dx and v = x. Using the formula for integration by parts, we get: ∫(x^2 + 16) dx = x(x^2 + 16) - ∫2x^2 dx Simplifying the integral on the right-hand side, we get: ∫(x^2 + 16) dx = x(x^2 + 16) - (2/3)x^3 + C where C is the constant of integration.

Now, let's substitute the limits of integration into the equation to find In: In = [ (22 +16) dx ] = ∫(x^2 + 16) dx evaluated from 2n to 2n+2 In = [(2n+2)((2n+2)^2 + 16) - (2n)((2n)^2 + 16)] - (2/3)[(2n+2)^3 - (2n)^3] Simplifying this expression, we get: In = 4n^3 + 24n^2 + 48n

Now, we need to find expressions for An and Bn such that In+1 = AnIn + Bn. Using the expression we just found for In, we can evaluate In+1 as: In+1 = 4(n+1)^3 + 24(n+1)^2 + 48(n+1) Expanding this expression, we get: In+1 = 4n^3 + 36n^2 + 124n + 144

Now, we can substitute In and In+1 into the equation In+1 = AnIn + Bn to get: 4n^3 + 36n^2 + 124n + 144 = A(n)(4n^3 + 24n^2 + 48n) + B(n) Simplifying this equation, we get: 4n^3 + 36n^2 + 124n + 144 = A(n)In + A(n)48n + B(n) Comparing coefficients, we get: A(n) = 4 B(n) = 36n^2 + 124n + 144

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The length of the 400-meter hurdle event completed by Jamal in feet is equals to 1312.0 feet approximately.

Conversion of meters to feet is equal to,

1 meter is approximately equal to 3.28 feet.

Length of the hurdles of events completed by Jamal on his track  = 400meters

So, the length of the race in feet can be calculated as,

1 meter = 3.28 feet

⇒ length of the race in feet = 400 meters ×  3.28 feet/meter

⇒ length of the race in feet  = 1312 feet

Rounding to the nearest tenth is equal to,

1312 feet ≈ 1312.0 feet

Therefore, the length of the race in feet is approximately 1312.0 feet.

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The probability that he will end up riding the bus is the complement of the probability that all 6 cars will take him into town, which is 1 - (1/3)^6.  So, the probability he will end up riding the bus is approximately 0.99981 or 99.981%.

Given that the professor's waiting time for the bus is uniformly (0,1), we need to find the probability that he gets a ride from a car before the bus arrives. Let's break it down step-by-step:

1. The waiting time for the bus is uniform on (0,1). This means the professor could wait anywhere between 0 and 1 hour for the bus, with equal probability.
The probability that the math professor will end up riding the bus can be found by calculating the probability that the waiting time for the next bus is less than the time it takes for 6 cars to pass by the bus stop.

Since the waiting time is uniformly distributed on (0,1), the probability that the waiting time is less than x is equal to x. Therefore, the probability that the waiting time is less than 6/60 (i.e. the time it takes for one car to pass by the bus stop) is 6/60 = 1/10.

The probability that one car will take him into town is 1/3, so the probability that all 6 cars will take him into town is (1/3)6.

2. Cars pass by at a rate of 6 per hour. Therefore, during the time the professor waits for the bus (0 to 1 hour), there will be 6 cars on average.

3. Each car will give the professor a ride with a probability of 1/3. So, the probability that a car won't give a ride is 2/3.

Now, let's calculate the probability that none of the 6 cars give the professor a ride:

(2/3)^6 = 0.08779 (approximately)

This is the probability that the professor won't get a ride from any of the 6 cars.

Since he either gets a ride from a car or takes the bus, the probability he will end up riding the bus is the complement of the probability he gets a ride from a car:

1 - 0.08779 = 0.91221 (approximately)

So, the probability the professor will end up riding the bus is approximately 0.91221, or 91.22%.

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Answer:

To factor the four-term polynomial 7x + 14 + xy + 2y by grouping, we can group the first two terms and the last two terms together as follows:

(7x + 14) + (xy + 2y)

We can factor 7 out of the first two terms and y out of the last two terms:

7(x + 2) + y(x + 2)

Now we can see that we have a common factor of (x + 2) in both terms. Factoring this out, we get:

(7 + y)(x + 2)

Therefore, the factored form of the polynomial 7x + 14 + xy + 2y is (7 + y)(x + 2).

The surface described by the equation φ=π/3 is a plane that intersects the sphere at a 60-degree angle.

In spherical coordinates, the angle φ represents the polar angle measured from the positive z-axis. When the polar angle is constant, the surface formed is a cone.

In this case, φ=π/3, which means the polar angle is always equal to π/3 (60 degrees). This results in a cone with its vertex at the origin, and it is symmetric about the positive z-axis. The cone has an opening angle of 2π/3 (120 degrees).

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The value is dy/dx = 2x / y. To find dy/dx using implicit differentiation, we differentiate both sides of the equation with respect to x:

d/dx(6x^2-3y^2) = d/dx(11)

Using the power rule for derivatives, we get:

12x - 6y(dy/dx) = 0

Now we can solve for dy/dx:

6y(dy/dx) = 12x

dy/dx = 2x/y

Therefore, the value of dy/dx for the given equation 6x^2-3y^2 = 11 is 2x/y.
Hi! I'd be happy to help you with implicit differentiation. Given the equation 6x^2 - 3y^2 = 11, we want to find dy/dx.

First, differentiate both sides of the equation with respect to x:

d/dx(6x^2) - d/dx(3y^2) = d/dx(11)

12x - 6y(dy/dx) = 0

Now, solve for dy/dx:

6y(dy/dx) = 12x

dy/dx = 12x / 6y

Your answer: dy/dx = 2x / y

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Answer: The unit price of the 120-fluid-ounce bottle of shampoo is $0.06 per fluid ounce.

Step-by-step explanation: To find the unit price of a 120-fluid-ounce bottle of shampoo that costs $7.20, we need to divide the total cost by the number of fluid ounces in the bottle.

Unit price = total cost/number of units

In this case, the total cost is $7.20 and the number of fluid ounces is 120. So the unit price is:

Unit price = $7.20 / 120 fluid ounces

Unit price = $0.06 per fluid ounce

Therefore, the unit price of the 120-fluid-ounce bottle of shampoo is $0.06 per fluid ounce.

To express the complex number -7i in the form R(cos(θ) + i sin(θ)) = Reil where R>0 and 0<θ<2π, we first need to find the magnitude R and the angle θ.The magnitude R of a complex number a+bi is given by |a+bi| = √(a^2 + b^2). In this case, a = 0 and b = -7, so |0-7i| = √(0^2 + (-7)^2) = 7. Therefore, R = 7.

The angle θ of a complex number a+bi is given by θ = atan(b/a) if a>0, θ = atan(b/a) + π if a<0 and b≥0, and θ = atan(b/a) - π if a<0 and b<0. In this case, a = 0 and b = -7, so θ = atan((-7)/0) + π = π/2.

Therefore, the complex number -7i can be expressed in the form R(cos(θ) + i sin(θ)) as 7(cos(π/2) + i sin(π/2)) = 7i(cos(0) + i sin(0)) = 7i, which can be written as Reil where R = 7, θ = π/2, and e^(iθ) = i.
To express the complex number -7i in the form R(cos(θ) + i sin(θ)) = Re^(iθ), follow these steps:

Step 1: Find the magnitude (R)
Since the complex number is -7i, its real part is 0 and its imaginary part is -7. Calculate the magnitude R using the formula:

R = √(Real part² + Imaginary part²) = √(0² + (-7)²) = √49 = 7

Step 2: Find the angle (θ)
Use the arctangent function to find the angle:

θ = arctan(Imaginary part / Real part) = arctan(-7 / 0)

Since the arctan function is not defined for division by zero, consider the quadrant of the complex number instead. In this case, -7i lies on the negative y-axis, which means the angle is:

θ = 270° or (3π/2 radians)

Step 3: Write the complex number in polar form
Now, write the complex number using R and θ:

-7i = 7(cos(3π/2) + i sin(3π/2)) = 7e^(i(3π/2))

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The definite integral I ≈ 0.006010 to six decimal places using the power series approximation.

To approximate the definite integral I = ∫0.4 ln(1+x^5) dx, we can use the power series expansion of ln(1+x) centered at x=0:

ln(1+x) = x - (x^2)/2 + (x^3)/3 - (x^4)/4 + ...

Substituting x^5 for x, we get:

ln(1+x^5) = x^5 - (x^10)/2 + (x^15)/3 - (x^20)/4 + ...

Integrating both sides from 0 to 0.4, we have:

I = ∫0.4 ln(1+x^5) dx

= ∫0.4 [x^5 - (x^10)/2 + (x^15)/3 - (x^20)/4 + ...] dx

= [x^(5+1)/(5+1)] - [(x^(10+1))/(2(10+1))] + [(x^(15+1))/(3(15+1))] - [(x^(20+1))/(4(20+1))] + ... | from 0 to 0.4

= [0.4^6/6] - [0.4^11/42] + [0.4^16/144] - [0.4^21/320] + ...

Using the first four terms of this series, we can approximate I to six decimal places as follows:

I ≈ [0.4^6/6] - [0.4^11/42] + [0.4^16/144] - [0.4^21/320]

≈ 0.006010

Therefore, the definite integral I ≈ 0.006010 to six decimal places using the power series approximation.

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Answer:

now

Step-by-step explanation:

ok the formula to convert your gpa into percentage is to just multiply your gpa by 25

Step-by-step explanation:

the slope of a line is defined by the factor "a" of x in an equation of the form y = ax + b

to be safe, let's transform

3x - y = 5

3x = y + 5

y = 3x - 5

the slope is 3, and any parallel line must have the same slope.

and for b we use the point coordinates :

-7 = 3×-3 + b

-7 = -9 + b

2 = b

the equation of the parallel line through (-3, -7) is

y = 3x + 2

In summary: sin(theta) = -55/143 and sin(2 theta) = -121/143 and cos(2 theta) = 14399/20449.

To find the exact value of sin(theta), we need to use the fact that cos(theta) = 132/143 and the terminal ray of theta is in Quadrant III. In this quadrant, the x-coordinate (cosine) is negative and the y-coordinate (sine) is also negative. So, we have:

sin^2(theta) = 1 - cos^2(theta)  (using the Pythagorean identity)
sin^2(theta) = 1 - (132/143)^2
sin^2(theta) = 1 - 17424/20449
sin^2(theta) = 3025/20449
sin(theta) = -55/143   (since sin(theta) is negative in Quadrant III)

Now, we need to find sin(2 theta). We can use the double angle identity:

sin(2 theta) = 2 sin(theta) cos(theta)

Plugging in the values we know, we get:

sin(2 theta) = 2 (-55/143) (132/143)
sin(2 theta) = -15840/20449

Finally, we need to find cos(2 theta). We can use the double angle identity:

cos(2 theta) = cos^2(theta) - sin^2(theta)

Plugging in the values we know, we get:

cos(2 theta) = (132/143)^2 - (-55/143)^2
cos(2 theta) = 17424/20449 - 3025/20449
cos(2 theta) = 14499/20449
Hi! Based on the given information, we have cos(theta) = 132/143, and theta lies in Quadrant III. We can use the Pythagorean identity sin^2(theta) + cos^2(theta) = 1 to find sin(theta):

sin^2(theta) = 1 - cos^2(theta)
sin^2(theta) = 1 - (132/143)^2
sin^2(theta) = 1 - 17424/20449
sin^2(theta) = 3025/20449

Since theta is in Quadrant III, sin(theta) will be negative. Therefore, sin(theta) = -sqrt(3025/20449) = -55/143.

Now, let's find sin(2 theta) and cos(2 theta) using the double-angle identities:

sin(2 theta) = 2 * sin(theta) * cos(theta)
sin(2 theta) = 2 * (-55/143) * (132/143)
sin(2 theta) = -121/143

cos(2 theta) = cos^2(theta) - sin^2(theta)
cos(2 theta) = (132/143)^2 - (-55/143)^2
cos(2 theta) = 17424/20449 - 3025/20449
cos(2 theta) = 14399/20449

In summary:
sin(theta) = -55/143
sin(2 theta) = -121/143
cos(2 theta) = 14399/20449

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The integral that gives the volume of the solid formed by revolving the region about the x-axis is V = [tex]\int\limits^{-2}_{-2}[/tex]π(4−x²)² dx is (8/3)π cubic units.

To find the volume of the solid formed by revolving the region about the x-axis, we can use the disk method.

First, we need to find the limits of integration. The given function y = 4 - x² intersects the x-axis at x = -2 and x = 2. So, the limits of integration will be from -2 to 2.

Next, we need to express the given function in terms of x. Solving for x, we get x = ±√(4-y).

Now, we can set up the integral for the volume using the disk method

V = π [tex]\int\limits^a_b[/tex] (f(x))² dx

where f(x) = √(4-x²), and a = -2, b = 2.

V = π [tex]\int\limits^{-2}_{-2}[/tex] (√(4-x²))² dx

V = π [tex]\int\limits^{-2}_{-2}[/tex] (4-x²) dx

V = π [4x - (1/3)x³] [tex]|^{-2}_2[/tex]

V = π [(32/3)-(8/3)]

V = (8/3)π

Therefore, the volume of the solid formed by revolving the region about the x-axis is (8/3)π cubic units.

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To find the values of trigonometric functions for 2θ, we'll need to use the double-angle identities.

Given that sin θ = 2/5 and θ lies in quadrant II, we can determine the values of the other trigonometric functions for θ using the Pythagorean identity: sin^2 θ + cos^2 θ = 1.

Let's start by finding cos θ:

sin θ = 2/5

cos^2 θ = 1 - sin^2 θ

cos^2 θ = 1 - (2/5)^2

cos^2 θ = 1 - 4/25

cos^2 θ = 21/25

Since θ lies in quadrant II, cos θ is negative:

cos θ = -√(21/25)

cos θ = -√21/5

Now, we can use the double-angle identities:

a. sin 2θ = 2sin θ cos θ

  sin 2θ = 2 * (2/5) * (-√21/5)

  sin 2θ = -4√21/25

b. cos 2θ = cos^2 θ - sin^2 θ

  cos 2θ = (21/25) - (4/25)

  cos 2θ = 17/25

c. tan 2θ = (2tan θ) / (1 - tan^2 θ)

  tan θ = sin θ / cos θ

  tan θ = (2/5) / (-√21/5)

  tan θ = -2√21/21

  tan 2θ = (2 * (-2√21/21)) / (1 - (-2√21/21)^2)

  tan 2θ = (-4√21/21) / (1 - (4(21)/21))

  tan 2θ = (-4√21/21) / (1 - 4)

  tan 2θ = (-4√21/21) / (-3)

  tan 2θ = 4√21/63

Therefore, the exact values for the given trigonometric functions are:

a. sin 2θ = -4√21/25

b. cos 2θ = 17/25

c. tan 2θ = 4√21/63

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Answer:

Step-by-step explanation:

( a ± b )² = a² ± 2ab + b²

a² - b² = ( a - b )( a + b )

~~~~~~~~~~~~~~~  

y = ( x + 3 )² - 25

(a). y = ( x² + 6x + 9 ) - 25

     y = x² + 6x - 16

(b). y = ( x + 3 )² - 25

     y = ( x + 3 )² - 5²

     y = [ ( x + 3 ) - 5 ] [ ( x + 3 ) + 5 ]

     y = ( x - 2 )( x + 8 )