kelvin makes a map of his apartment using a coordinate system with yards as the units. the point (-9, 8) represents the main entrance of the apartment and the point (-5, 6) represents the location of the kitchen. approximately how far apart are the main entrance and the kitchen?

The area of the region that lies inside both curves is 27/10 square units.

To find the area of the region that lies inside both curves, we need to set the equations equal to each other and find the points of intersection:

3 cos(theta) = sin(theta)

Dividing both sides by cos(theta) (since cos(theta) cannot be zero), we get:

3 = tan(theta)

Taking the arctangent of both sides, we get:

theta = arctan(3)

So the curves intersect at theta = arctan(3).

To find the area of the region, we integrate the equation for the smaller curve (r = sin(theta)) squared from 0 to arctan(3), and subtract the integral of the equation for the larger curve (r = 3cos(theta)) squared from 0 to arctan(3):

[tex]Area = ∫[0, arctan(3)] (sin(theta))^2 d(theta) - ∫[0, arctan(3)] (3cos(theta))^2 d(theta)[/tex]

Simplifying and evaluating the integrals, we get:

Area = (9/4)sin(2arctan(3)) - 27/2

Using the trigonometric identity [tex]sin(2arctan(x)) = (2x)/(1+x^2),[/tex] we get:

Area = (27/5) - 27/2

Area = 27/10

Therefore, the area of the region that lies inside both curves is 27/10 square units.

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There is nothing inherently wrong with any of the statements in question 21. Sampling rate and sound frequency both use hertz (Hz) as their unit of measurement.

The sample rate is a setting that can be adjusted during the digitization process, as is the sound frequency. It is also true that a higher sampling rate does not necessarily result in a higher pitch.

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A column chart is an excellent tool for comparing values side by side, broken down by category. With its clear and concise display of data, it is an invaluable asset for businesses, researchers, and anyone looking to understand complex information quickly and easily.

To compare values side by side, broken down by category, a column chart is an effective tool. Column charts are ideal for comparing data across different categories or groups, as they clearly display the differences between values. When creating a column chart, the categories are listed on the horizontal axis, and the values are listed on the vertical axis. Each column represents a different category, and the height of the column corresponds to the value of that category. Column charts can be customized to fit specific needs, such as adding colors or labels to each category. They are also versatile and can be used to display a wide range of data, from sales figures to survey results. You should use a bar chart to compare values side by side, broken down by category. Bar charts are a helpful visualization tool that allows you to compare data across different categories in an easy-to-read format.

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The volume of the box with surface area 10 is given by the formula V = [tex]2.5x^2 - 0.25x^4,[/tex] where x is the length of a side of the square base.

To find a formula for the volume of the box with surface area A and square base with side x, we first need to find the height of the box. Since the box has a square base, the area of the base is [tex]x^2[/tex]. The remaining surface area is the sum of the areas of the four sides, each of which is a rectangle with base x and height h. Therefore, the surface area A is given by:

A = [tex]x^2 + 4xh[/tex]

Solving for h, we get:

h = [tex](A - x^2) / 4x[/tex]

The volume V of the box is given by:

V = [tex]x^2 * h[/tex]

The domain of V is all non-negative real numbers, since both [tex]x^2[/tex] and A are non-negative.

V as a function of x, we can use a graphing calculator or plot points using a table of values. The graph will be a parabola opening downwards, with x-intercepts at 0 and (A) and a maximum at x = sqrt(A) / sqrt(2).

To find the maximum value of V, we can take the derivative of V with respect to x and set it equal to 0:

dV/dx =[tex](2Ax - 4x^3) / 4[/tex]

Setting this equal to 0 and solving for x, we get:

To find the formula for the volume of a box having surface area 10, we simply replace A with 10 in the formula we derived earlier:

V =[tex](10x^2 - x^4) / 4[/tex]

Simplifying, we get:

V = [tex]2.5x^2 - 0.25x^4[/tex]

Therefore, the volume of the box with surface area 10 is given by the formula V = [tex]2.5x^2 - 0.25x^4[/tex], where x is the length of a side of the square base. The domain of V is all non-negative real numbers.

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Correct Question:

Example 4 A closed box has a fixed surface area A and a square base with side x. (a) Find a formula for the volume, V. of the box as a function of x. What is the domain of V? (b) Graph V as a function of x. (c) Find the maximum value of V.

use the work in example 4 in this section of the textbook to find a formula for the volume of a box having surface area 10.

Answer:

Step-by-step explanation:

We can start by setting up equations for the height of each snowman as a function of time. Let t be the time in hours after sunrise.

For Snowman A, the height as a function of time is given by:

A(t) = 36 - 3t

For Snowman B, the height as a function of time is given by:

B(t) = 57 - 6t

To find when the two snowmen are the same height, we can set the two equations equal to each other and solve for t:

36 - 3t = 57 - 6t

3t = 21

t = 7

So the two snowmen will be the same height after 7 hours.

To find the height of each snowman at that time, we can substitute t = 7 into the equations:

A(7) = 36 - 3(7) = 15 inches

B(7) = 57 - 6(7) = 15 inches

Therefore, both Snowman A and Snowman B will be 15 inches tall after 7 hours.

To graph the functions, we can plot points for various values of t and connect them with a straight line:

For A(t):

t | A(t)

--|-----

0 | 36

1 | 33

2 | 30

3 | 27

4 | 24

5 | 21

6 | 18

7 | 15

For B(t):

t | B(t)

--|-----

0 | 57

1 | 51

2 | 45

3 | 39

4 | 33

5 | 27

6 | 21

7 | 15

The graph of both functions is a straight line with a negative slope. The two lines intersect at (7, 15), which represents the point in time when both snowmen are the same height.

The next number in sequence 6 3 12 9 36 33 is 132.

Sequence is an enumerated collection of objects in which repetitions are allowed and order matters.

The pattern in the sequence of numbers seems to be:

6-3 =3

3×4=12

12-3=9

9×4=36

36-3=33

33×4=132

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The probability that a randomly selected graduate will have a student loan greater than $30,000 is 0.1587, the probability that the loan is less than $22,500 is 0.3085, and the probability that the loan falls between $20,000 and $32,000 is 0.8186.

Let X be a random variable representing the student loans of graduates. Then, X ~ N(μ = 25,000, σ = 5,000). To find the probabilities, we need to standardize the values using the standard normal distribution, Z ~ N(0, 1), where Z = (X - μ) / σ.

a. P(X > 30,000) = P(Z > (30,000 - 25,000) / 5,000) = P(Z > 1) = 0.1587

b. P(X < 22,500) = P(Z < (22,500 - 25,000) / 5,000) = P(Z < -0.5) = 0.3085

c. P(20,000 < X < 32,000) = P((20,000 - 25,000) / 5,000 < Z < (32,000 - 25,000) / 5,000) = P(-1 < Z < 1.4) = 0.8186

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If in a certain town, the mean price of a chocolate chip cookie is $2.30 and the standard deviation is $0.40 at various bakeries. The probability is: 0.986.

Let x represent  the price of a chocolate chip cookie

Let y  represent the  price of a brownie.

Let assume x ~ n(2.30, 0.40)

y ~ n(3.50, 0.20)

x and y are independent

Let t represent the  total amount of money

t = 7x + 3y

Mean and standard deviation of t is  

= 7(2.30) + 3(3.50)

= 26.6

Variance = 7^2(0.40)^2 + 3^2(0.20)^2

= 8.2

So, t ~ n is (23.1, √8.2)

Now let find the  probability

Z = (28 - 23.1) / √2.45

Z=  3.13

Using a standard normal table

Probability = 0.986.

Therefore the probability  is 0.986.

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1) The total amount of wrapper needed is = 325.2 inches2

2) The amount of wrapper needed is 180.8 in². The wrapper remaining would be 419.2in²

3) the amount of canvas fabric required to make the tent including the floor is 104.6ft²

4) total suface area of the square pyramid is 62.62cm²

1)  Surface Area = 2×(9×12 + 9×2.6 + 12×2.6) =325.2 inches2

2) Surface Area = 2×(10×7 + 10×1.2 + 7×1.2) = 180.8 inches2

3)

3(l x w)

2(1/2 (bh)

L = 6ft

W = 4.7ft

⇒ 3 (6 x 4.7)

= 84.6

2(1/2 (bh))

⇒ 2 (1/2 (4 x 5)

=  20

Thus total surface area = 20 + 84.6

= 104.6ft²

4) For this case, there are 4 triangles and one square base.

Thus total surface area =

4( 1/2 (bh) + (l²)

⇒ 4 (1/2 (3.9 x 3.1) + (3.1²)

= 62.62cm²

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(A) The percentage is approximately 62.07%.

(B) percentage of years will have an annual rainfall of more than 39 inches is 72.17%.

(C) the smaller percentage from the larger one: 50.48% - 22.17% = 28.31%.

First, let's recall that the normal distribution is characterized by the mean and standard deviation. In this case, the mean annual rainfall is 42.3 inches and the standard deviation is 5.6 inches.

To answer your questions, we'll use the Z-score formula: Z = (X - mean) / standard deviation. Then, we can use a Z-table or calculator to find the percentage.

a) For annual rainfall less than 44 inches:
Z = (44 - 42.3) / 5.6 = 1.7 / 5.6 ≈ 0.3036
Using a Z-table or calculator, the percentage is approximately 62.07%.

b) For annual rainfall more than 39 inches:
Z = (39 - 42.3) / 5.6 = -3.3 / 5.6 ≈ -0.5893
Using a Z-table or calculator, the percentage for LESS than 39 inches is approximately 27.83%. To find the percentage of years with more than 39 inches, subtract from 100%: 100% - 27.83% = 72.17%.

c) For annual rainfall between 38 and 43 inches:
Z1 = (38 - 42.3) / 5.6 ≈ -0.7679
Z2 = (43 - 42.3) / 5.6 ≈ 0.1250
Using a Z-table or calculator, the percentage for Z1 is 22.17%, and for Z2 is 50.48%. To find the percentage between these two Z-scores, subtract the smaller percentage from the larger one: 50.48% - 22.17% = 28.31%.

So, the answers are: a) 62.07%, b) 72.17%, and c) 28.31%.

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The centroid of the region bounded by the curves y = 6√x and y = 2x is (3.6,0.5).

To find the centroid of the region bounded by the curves y = 6√x and y = 2x, we first need to find the limits of integration.

Since y = 6√x and y = 2x intersect at y = 0, we can set the two equations equal to each other to find where they intersect:

6√x = 2x

36x = 4x²

x² - 9x = 0

x(x - 9) = 0

Therefore, the curves intersect at x = 0 and x = 9.

Next, we need to set up the integrals for the x-coordinate and y-coordinate of the centroid:

x-bar = [tex]\frac{1}{A} \int_a^bxf(x)dx[/tex]

(1/A) * [tex]\int_a^b[/tex] x*f(x) dx

y-bar = [tex]\frac{1}{A} \int_a^b\frac{1}{2} (f(x))^2dx[/tex]

where f(x) is the distance between the two curves at x, and A is the area of the region bounded by the curves.

The distance between the two curves at x is:

f(x) = 6√x - 2x

The area of the region is:

A = [tex]\int_0^9[/tex] (6√x - 2x) dx

Evaluating this integral, we get:

A = 27

Now we can find the x-coordinate of the centroid:

x-bar = [tex]\frac{1}{27} \int_0^9x(6\sqrt{x} -2x)dx[/tex]

Simplifying and evaluating this integral, we get:

x-bar = 3.6

The y-coordinate of the centroid:

y-bar = [tex]\frac{1}{27} \int_0^9\frac{1}{2} (6\sqrt{x} - 2x)^2dx[/tex]

Simplifying and evaluating this integral, we get:

y-bar = 0.5

Therefore, the centroid of the region bounded by the curves y = 6√x and y = 2x is (3.6,0.5).

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To solve the homogeneous differential equation:

dy/dx = (y - x)/(y + x)

We can make the substitution y = vx, where v is a new variable.

Differentiating both sides of the substitution with respect to x:

dy/dx = v + x * dv/dx

Now we substitute the new variables into the original differential equation:

v + x * dv/dx = (vx - x)/(vx + x)

Next, we simplify the equation:

v + x * dv/dx = (v - 1)/(v + 1)

To separate variables, we move the terms involving v to one side and the terms involving x to the other side:

(v + 1) * dv/(v - 1) = -x * dx

Now we can integrate both sides:

∫(v + 1)/(v - 1) * dv = -∫x dx

To integrate the left side, we use partial fractions:

∫(v + 1)/(v - 1) * dv = ∫(1 + 2/(v - 1)) dv

∫(v + 1)/(v - 1) * dv = v + 2ln|v - 1| + C1

For the right side, we integrate:

-∫x dx = -0.5x^2 + C2

Putting it all together:

v + 2ln|v - 1| = -0.5x^2 + C

Substituting back y = vx:

y + 2ln|y - x| = -0.5x^2 + C

This is the general solution to the homogeneous differential equation.

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f(n) = (n-1)!/2^(n-2) for n > 1.

To find the function f(n), we need to use the given information that ai = 1 and an+1 (n+1)an 2 for n > 1. We can use mathematical induction to derive a formula for An.

First, we can find A2:

a3 = 3a2/2
a2 = 2a3/3

Substituting a2 in terms of a3, we get:

2a3/3 = 1
a3 = 3/2

Thus, A2 = 2a3/3 = 1.

Next, we assume that An = f(n) for some function f, and we want to find a formula for An+1. Using the given relation, we have:

An+1 = (n+1)An/2

Substituting f(n) for An, we get:

f(n+1) = (n+1)f(n)/2

Now, we can use this recursive formula to find f(n) for all n > 1. Starting with f(2) = 1, we can apply the formula repeatedly:

f(3) = 3/2
f(4) = 3/4
f(5) = 15/16
f(6) = 45/32
f(7) = 315/64
...

Thus, the function f(n) is:

f(n) = (n-1)!/2^(n-2) for n > 1.

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a. There should be  400/3 units of the basic model and 200/3 units of the luxury model sold per week in order to maximize the company's total weekly revenue.

b. The maximum value of the total weekly revenue is -25/2.

Part (a):

To maximize the total weekly revenue, we need to find the critical points of the revenue function R(x,y), where the partial derivatives are zero or do not exist.

∂R/∂x = 160 - 0.4x - 0.2y = 0 ...... (1)

∂R/∂y = 220 - 0.2x - 0.3y = 0 ...... (2)

Solving these two equations simultaneously, we get:

x = 400/3 and y = 200/3

Substituting these values of x and y into the revenue function R(x,y), we get:

R(400/3, 200/3) = $70,266.67

Therefore, the company should sell 400/3 units of the basic model and 200/3 units of the luxury model per week to maximize the total weekly revenue.

Part (b):

To find the maximum value of the total weekly revenue, we need to evaluate the revenue function R(x,y) at the critical point (400/3, 200/3) and at the endpoints of the feasible region (where x and y are non-negative).

At (400/3, 200/3), we have:

R(400/3, 200/3) = $70,266.67

At the endpoints of the feasible region, we have:

R(0,0) = $0

R(0,1466.67) = $32,133.33

R(2666.67,0) = $42,666.67

Therefore, the maximum value of the total weekly revenue is $70,266.67 when the company sells 400/3 units of the basic model and 200/3 units of the luxury model per week.

Example 1:

To find the relative extrema of the function f(x,y) = x^2 + y^2 - 9x - 7y, we need to find the critical points of the function, where the partial derivatives are zero or do not exist.

∂f/∂x = 2x - 9 = 0 ...... (1)

∂f/∂y = 2y - 7 = 0 ...... (2)

Solving these two equations simultaneously, we get:

x = 9/2 and y = 7/2

Substituting these values of x and y into the function f(x,y), we get:

f(9/2, 7/2) = -25/2

Therefore, the critical point (9/2, 7/2) is a relative maximum of the function f(x,y), and the maximum value is -25/2.

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To describe y as the sum of two orthogonal vectors, x1 in the span{u} and x2 orthogonal to u , we follow two steps procedure:

1.First, find a vector x1 in the span{u} that is the projection of y onto u. To do this, use the formula:
  x1 = (y • u / ||u||^2) * u, where • represents the dot product and || || represents the magnitude of the vector.

2.Next, find the vector x2 that is orthogonal to u. Since y can be represented as the sum of x1 and x2, you can find x2 by subtracting x1 from y:
  x2 = y - x1

3.Now, you have y as the sum of two orthogonal vectors x1 and x2, with x1 in the span{u} and x2 orthogonal to u.

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So, x must be an odd multiple of 1/2 in a triangle. The largest odd possible value of x in multiple of 1/2 and less than 50 is 4x + 2 .

Therefore, the largest possible value of x is 49/2. we know that 4x + 2 is even since 2 is even and 4x is even for any integer x. Therefore, x must be a multiple of 1/2 for 4x + 2 to be an integer.

Here we can set up the following equations:

PR = 2PQ + 12

QR = 4PQ + 2

Substituting PQ = x into these equations, we get:

PR = 2x + 12

QR = 4x + 2

For the triangle to have integer side lengths, PR, PQ, and QR must all be integers.

We know that 2x + 12 is even since 12 is even and 2x is even for any integer x.

Therefore, x must be odd for 2x + 12 to be an integer.

Similarly, we know that 4x + 2 is even since 2 is even and 4x is even for any integer x.

Therefore, x must be a multiple of 1/2 for 4x + 2 to be an integer.( from figure we get 4x+2).

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Correct Question:

In the given figure, PR is 12 more than twice PQ, and QR is two more than four times PQ. If all three sides of the triangle have integer lengths, what is the largest possible value of x?

To find f(g(x)), we need to first evaluate g(x) and then substitute the result in f(x). Therefore, we have: f(g(x)) = f(x + 6) = -5(x + 6) = -5x - 30.

To find g(f(x)), we need to first evaluate f(x) and then substitute the result in g(x). Therefore, we have:

g(f(x)) = g(-5x) = -5x + 6

To find f(f(x)), we need to substitute f(x) into the expression for f(x). Therefore, we have:

f(f(x)) = f(-5x) = -5(-5x) = 25x

Therefore, the answers are:

a. f(g(x)) = -5x - 30

b. g(f(x)) = -5x + 6

c. f(f(x)) = 25x

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Chebyshev's inequality states that for any random variable X with finite mean μ and finite variance σ^2, and for any positive value k, the probability that X deviates from its mean by more than k standard deviations is at most [tex]1/k^2[/tex]:

P(|X - μ| ≥ kσ) ≤ 1/[tex]k^2[/tex]

In this case, X ~ Poisson(λ), which means that the mean and variance of X are both equal to λ.

Applying Chebyshev's inequality with k = √λ gives:

P(|X - λ| ≥ √λ √λ) ≤ 1/λ

which simplifies to:

P(|X - λ| ≥ λ) ≤ 1/λ

Now, we want to show that X/λ → 1 as λ → ∞. This is equivalent to showing that:

lim λ→∞ P(|X/λ - 1| ≥ ε) = 0 for any ε > 0.

We can rewrite this as:

lim λ→∞ P(|X - λ| ≥ ελ) = 0

So, we need to choose k = ελ/√λ = ε√λ. Then, we have:

P(|X - λ| ≥ ελ) ≤ P(|X - λ| ≥ ε√λ√λ) ≤ 1/(ε^2λ)

Taking the limit as λ → ∞, we get:

lim λ→∞ P(|X - λ| ≥ ελ) ≤ lim λ→∞ 1/(ε^2λ) = 0

Therefore, we have shown that X/λ → 1 as λ → ∞, as required.

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The absolute value function as a piecewise function is

Given that

f(x) = |-x^2 + 2x + 8|

When the expression is factored, we have

f(x) = |-(x + 2)(x - 4)|

Set the expression in the absolute bracket to 0

This gives

-(x + 2)(x - 4) = 0

When the equation is solved for x, we have

x = -2 and x = 4

These values represent the boundaries of the piecewise function

So, we have

f(x) = -(-x^2 + 2x + 8), x < -2 and x > 4

f(x) = -x^2 + 2x + 8, -2 ≤ x ≤ 4

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A new college graduate in business would need to earn at least $78,278 in order to have a starting salary higher than % of starting salaries of new college graduates in health sciences.

a. To find the probability that a new college graduate in business will earn a starting salary of at least X, we need to calculate the z-score:
z = (X - ) /
Using the given values, we have:
z = (X - ) /
z = (X - ) /
From the z-table, we can find the probability corresponding to this z-score. For example, if X = 60,000, then:
z = (60,000 - ) /
z = (60,000 - ) /
Looking up this z-score in the table, we find the probability to be approximately 0.8643. Therefore, the probability that a new college graduate in business will earn a starting salary of at least $60,000 is 0.8643.
b. Similarly, to find the probability that a new college graduate in health sciences will earn a starting salary of at least Y, we need to calculate the z-score:
z = (Y - ) /
Using the given values, we have:
z = (Y - ) /
z = (Y - ) /
From the z-table, we can find the probability corresponding to this z-score. For example, if Y = 50,000, then:
z = (50,000 - ) /
z = (50,000 - ) /
Looking up this z-score in the table, we find the probability to be approximately 0.9332. Therefore, the probability that a new college graduate in health sciences will earn a starting salary of at least $50,000 is 0.9332.
c. To find the probability that a new college graduate in health sciences will earn a starting salary less than Z, we need to calculate the z-score:
z = (Z - ) /
Using the given values, we have:
z = (Z - ) /
z = (Z - ) /
From the z-table, we can find the probability corresponding to this z-score. For example, if Z = 45,000, then:
z = (45,000 - ) /
z = (45,000 - ) /
Looking up this z-score in the table, we find the probability to be approximately 0.8289. Therefore, the probability that a new college graduate in health sciences will earn a starting salary less than $45,000 is 0.8289.
d. To find the salary that a new college graduate in business would need to earn in order to have a starting salary higher than % of starting salaries of new college graduates in health sciences, we need to find the z-score corresponding to this percentile.
From the given information, we know that  is the average starting salary for new college graduates in health sciences, and  is the standard deviation. Using the z-table, we can find the z-score corresponding to the percentile. For example, if we want to find the salary that is higher than % of starting salaries of new college graduates in health sciences, then the z-score is approximately 1.44.
Now, we can use the z-score formula to find the corresponding salary for a new college graduate in business:
z = (X - ) /
1.44 = (X - ) /
Solving for X, we get:
X =  + 1.44
Using the given values, we have:
X =  + 1.44
X =  + (1.44 x )
Substituting the values of  and , we get:
X =  + (1.44 x )
X =  + (1.44 x )
Rounding to the nearest whole number, we get:
X = $78,278

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Answer: 2/3

Step-by-step explanation:

frequency, or the variable b, is related to the period. (Specifically, it frequency is the value you divide the original period by to get the new period).

the original unchanged period of the tangent function is pi.

In this case however, we see the period, (one cycle), is going from 3pi to 9pi/2. this means the period is 3pi/2. Since we divide pi by 2/3 to get to 3pi/2, our frequency, or b, is 2/3.

1. The value of b is 0 when the angle between v = (b, 2) and w = (-8,-6) is π/4

2. A unit vector perpendicular to the plane = (1/√3, -1/√3, 1/√3)

3.  The equation of the plane containing the line x = -1+t, y = 1 – 2t, z=t    6x + 6y - 18z + 36 = 0

a) Find th value of b when the angle between v = (b, 2) and w = (-8,-6) is π/4.

                        tanθ = (y2 - y1) / (x2 - x1)

                          θ = π/4

                           tanπ/4 = 1

1 = (-6 - 2) / (-8 - b)

1 = -8 / (-8 - b)

-8 - b = 8

b = -16

(b) PQ = Q - P = (-1 - 2, 1 - 1, 2 - (-1)) = (-3, 0, 3)

PR = R - P = (1 - 2, -1 - 1, 2 - (-1)) = (-1, -2, 3)  

PQ x PR = (0 x 3 - (-2) x 3, (-3) x 3 - (-1) x 3, (-3) x (-2) - 0 x (-1)) = (6, -6, 6)

||PQ x PR|| =√(6² + (-6)² + 6²) =

√(36 + 36 + 36)

=√108

= 6√3

Unit vector perpendicular to the plane

= (6 / (6√3), -6 / (6√3), 6 / (6√3)

= (1/√3, -1/√3, 1/√3)

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The product (64)(81)(125) has 140 different positive integer factors.

To find the number of positive integer factors of the product (64)(81)(125), we first need to determine the prime factorization of each number.

64 = 2^6 (as it is 2 multiplied by itself 6 times)
81 = 3^4 (as it is 3 multiplied by itself 4 times)
125 = 5^3 (as it is 5 multiplied by itself 3 times)

Now, let's consider the product (64)(81)(125) = (2^6)(3^4)(5^3). To find the number of different positive integer factors, we use the formula:

(Number of factors of the first prime + 1) * (Number of factors of the second prime + 1) * (Number of factors of the third prime + 1)

In our case, the formula would be:

(6 + 1) * (4 + 1) * (3 + 1) = 7 * 5 * 4

Calculating the result, we get:

7 * 5 * 4 = 140

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The values of c that make f continuous everywhere: ae +4 if I <0 f(x) cx + 5 if OSI 31 if 1 are c = 26 and a = 1.

To find the values of c that make f continuous everywhere, we need to ensure that the left and right limits at x = 0 and x = 1 are equal.

Starting with x < 0: - The function f(x) = ae + 4 for x < 0. Next, consider x > 1: - The function f(x) = 31 for x > 1. Finally, for 0 < x < 1: - The function f(x) = cx + 5.

To make f continuous everywhere, we need to make sure that the value of f(x) from both sides of x = 0 and x = 1 are equal.

For x = 0: - The left limit of f(x) is ae + 4. - The right limit of f(x) is c(0) + 5 = 5. For these limits to be equal, ae + 4 = 5.

Solving for a, we get a = 1. For x = 1: - The left limit of f(x) is c(1-) + 5. - The right limit of f(x) is 31. For these limits to be equal, we need to make sure that c(1-) + 5 = 31. Solving for c, we get c = 26.

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The limits of integration for θ are: 0 ≤ θ ≤ 2π

To find the volume of the region between the paraboloid and the cone, we first need to determine the limits of integration. We will use cylindrical coordinates to solve this problem.

The cone is given by the equation[tex]z = 2s(x^2 + y^2)^0.5[/tex], which in cylindrical coordinates becomes z = 2sr. The paraboloid is given by the equation [tex]z = 24 - 2r^2[/tex], where [tex]r^2 = x^2 + y^2[/tex].

The intersection of the paraboloid and the cone occurs where:

[tex]24 - 2r^2 = 2sr2r^2 + 2sr - 24 = 0r^2 + sr - 12 = 0[/tex]

Using the quadratic formula, we find that:

[tex]r = (-s ± (s^2 + 48)^0.5)/2[/tex]

Since r must be positive, we take the positive root:

[tex]r = (-s + (s^2 + 48)^0.5)/2[/tex]

The limits of integration for s are then:

[tex]0 ≤ s ≤ (48)^0.5[/tex]

The limits of integration for r are:

[tex]0 ≤ r ≤ (-s + (s^2 + 48)^0.5)/2[/tex]

The limits of integration for θ are:

0 ≤ θ ≤ 2π

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Answer: 72 students liked PT or DH.

To determine the number of students who liked Pizza Tower (PT) or Dining Hall (DH), we can use the principle of inclusion-exclusion.

Given the following information:

- 27 liked AL but not PT (AL - PT)

- 13 liked AL only (AL)

- 43 liked AL (AL)

- 41 liked PT (PT)

- 59 liked DH (DH)

- 13 liked PT and AL but not DH (PT ∩ AL - DH)

- Unknown: Number of students who liked PT or DH (PT ∪ DH)

We can use the formula:

n(A ∪ B) = n(A) + n(B) - n(A ∩ B)

Let's calculate the number of students who liked PT or DH:

n(PT ∪ DH) = n(PT) + n(DH) - n(PT ∩ DH)

We are given that 59 students liked DH, and 41 students liked PT. However, we need to determine the number of students who liked both PT and DH (n(PT ∩ DH)).

Using the principle of inclusion-exclusion, we have the following information:

- 13 liked PT and AL but not DH (PT ∩ AL - DH)

- 59 liked DH (DH)

- 13 liked PT and AL but not DH (PT ∩ AL - DH)

To find n(PT ∩ DH), we subtract the number of students who liked PT and AL but not DH from the total number who liked PT (PT):

n(PT ∩ DH) = n(PT) - n(PT ∩ AL - DH)

n(PT ∩ DH) = 41 - 13 = 28

Now, we can calculate the number of students who liked PT or DH:

n(PT ∪ DH) = n(PT) + n(DH) - n(PT ∩ DH)

n(PT ∪ DH) = 41 + 59 - 28

n(PT ∪ DH) = 72

Therefore, 72 students liked PT or DH.

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a. Domain: All real numbers

b. Intercepts: x-intercepts at (0,0), (6,0), and y-intercept at (0,0)

c. Symmetry: None

d. Asymptotes: None

e. Intervals of increase or decrease: Decreases on (-∞, 2), increases on (2, 6), and decreases on (6, ∞)

f. Local maximum or minimum values & Concavity and points of inflection: Local minimum at (2,-16) and point of inflection at (4,8), concave up on (2, 4) and concave down on (4, ∞)

h. Sketch the curve: See attached image.

a. Domain: The domain of a polynomial function is all real numbers. Therefore, the domain of f(x) = x^3 - 12x^2 + 36x is (-∞, ∞).

b. Intercepts: To find the x-intercepts, we set y = 0 and solve for x. Thus, we get x(x-6)(x-6) = 0 which gives x = 0 and x = 6 (multiplicity 2). The y-intercept is f(0) = 0.

c. Symmetry: We check for symmetry by replacing x with -x and simplifying. We get f(-x) = -x^3 - 12x^2 - 36x = -f(x), which means the function is not symmetric about the y-axis or origin.

d. Asymptotes: As the degree of the polynomial is 3, there are no horizontal or vertical asymptotes.

e. Intervals of increase or decrease: To find the intervals of increase or decrease, we take the first derivative and solve for critical points. f'(x) = 3x^2 - 24x + 36 = 3(x-2)(x-6). This gives critical points at x = 2 and x = 6. Thus, f(x) is decreasing on (-∞, 2) and (6, ∞) and increasing on (2, 6).

f. Local maximum or minimum values & Concavity and points of inflection: To find local maxima and minima, we take the second derivative and evaluate at critical points. f''(x) = 6x - 24. At x = 2, f''(2) = -12 which means there is a local minimum at (2,-16). To find points of inflection, we set f''(x) = 0 and solve for x. We get x = 4.

Thus, there is a point of inflection at (4,8). To determine concavity, we look at the sign of f''(x). f''(x) is negative on (2, 4) which means the function is concave down and f''(x) is positive on (4, ∞) which means the function is concave up.

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The total amount of cheese used in all the recipes is 29 3/4 cups.

Given is line plot we need to find the amount of cheese used in all the recipe,

so, 3 1/2 cups are used for one time, 3 3/4 used for 6 times and 4 cups are used for 1.

So,

3 1/4 + 3 3/4 × 6 + 4

= 13/4 + 15/4 × 6 + 4

= (13 + 90 + 16) / 4

= 119 / 4

= 29 3/4

Hence, the total amount of cheese used in all the recipes is 29 3/4 cups.

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