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2 years ago

8

ou are having a meeting with the CEO of a technology company. You have interpreted the number of laptops produced versus profit

as the function P(x) = x4 − 3x3 − 8x2 + 12x + 16. Describe to the CEO what the graph looks like and, in general, how to sketch the graph without using technology. Use complete sentences, and focus on the end behaviors of the graph and where the company will break even (where P(x) = 0).

1 answer:

Mama L [17]2 years ago

4 0

Answer: We can plot the graph with help of below explanation.

Step-by-step explanation:

Since, given equation of polynomial,

$P(x) = x^4 - 3x^3 - 8x^2 + 12x + 16$

End behavior : Since, the leading coefficient of the polynomial is positive and even.

Therefore, the end behavior of the polynomial is,

$f(x)\rightarrow -\infty$ as $x\rightarrow -\infty$

And, $f(x)\rightarrow +\infty$ as $x\rightarrow +\infty$

Points of the curve : since, P(4) = 0

Therefore, (x-4) is the multiple of P(x),

And we can write, $x^4 - 3x^3 - 8x^2 + 12x + 16= (x-4)(x^3+x^2-4x-4)$

$x^4 - 3x^3 - 8x^2 + 12x + 16=(x-4)(x+1)(x^2-4)$

$x^4 - 3x^3 - 8x^2 + 12x + 16= (x-4)(x+1)(x+2)(x-2)$

Thus, the roots of equation are 4, 2, -1 and -2.

Therefore, x-intercepts of the polynomial are (4,0) (2,0) (-1,0) and (-2,0)

Also, the y-intercept of the polynomial is ( 0,16)

Thus, we can plot the graph with help of the above information.

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Let D be the smaller cap cut from a solid ball of radius 8 units by a plane 4 units from the center of the sphere. Express the v

natima [27]

Answer:

Step-by-step explanation:

The equation of the sphere, centered a the origin is given by $x^2+y^2+z^2 = 64$ . Then, when z=4, we get

$x^2+y^2= 64-16 = 48$ .

This equation corresponds to a circle of radius $4\sqrt[]{3}$ in the x-y plane

c) We will use the previous analysis to define the limits in cartesian and polar coordinates. At first, we now that x varies from $-4\sqrt[]{3}$ up to $4\sqrt[]{3}$ . This is by taking y =0 and seeing the furthest points of x that lay on the circle. Then, we know that y varies from $-\sqrt[]{48-x^2}$ and $\sqrt[]{48-x^2}$ , this is again because y must lie in the interior of the circle we found. Finally, we know that z goes from 4 up to the sphere, that is , z goes from 4 up to $\sqrt[]{64-x^2-y^2}$

Then, the triple integral that gives us the volume of D in cartesian coordinates is

$\int_{-4\sqrt[]{3}}^{4\sqrt[]{3}}\int_{-\sqrt[]{48-x^2}}^{\sqrt[]{48-x^2}} \int_{4}^{\sqrt[]{64-x^2-y^2}} dz dy dx$ .

b) Recall that the cylindrical coordinates are given by $x=r\cos \theta, y = r\sin \theta,z = z$ , where r corresponds to the distance of the projection onto the x-y plane to the origin. REcall that $x^2+y^2 = r^2$ . WE will find the new limits for each of the new coordinates. NOte that, we got a previous restriction of a circle, so, since $\theta[\tex] is the angle between the projection to the x-y plane and the x axis, in order for us to cover the whole circle, we need that [tex]\theta$ goes from 0 to $2\pi$ . Also, note that r goes from the origin up to the border of the circle, where r has a value of 4\sqrt[]{3}. Finally, note that Z goes from the plane z=4 up to the sphere itself, where the restriction is \sqrt[]{64-r^2}. So, the following is the integral that gives the wanted volume

$\int_{0}^{2\pi}\int_{0}^{4\sqrt[]{3}} \int_{4}^{\sqrt[]{64-r^2}} rdz dr d\theta$ . Recall that the r factor appears because it is the jacobian associated to the change of variable from cartesian coordinates to polar coordinates. This guarantees us that the integral has the same value. (The explanation on how to compute the jacobian is beyond the scope of this answer).

a) For the spherical coordinates, recall that $z = \rho \cos \phi, y = \rho \sin \phi \sin \theta, x = \rho \sin \phi \cos \theta$ . where $\phi$ is the angle of the vector with the z axis, which varies from 0 up to pi. Note that when z=4, that angle is constant over the boundary of the circle we found previously. On that circle. Let us calculate the angle by taking a point on the circle and using the formula of the angle between two vectors. If z=4 and x=0, then y=4 $\sqrt[]{3}$ if we take the positive square root of 48. So, let us calculate the angle between the vector $a=(0,4\sqrt[]{3},4)$ and the vector b =(0,0,1) which corresponds to the unit vector over the z axis. Let us use the following formula

$\cos \phi = \frac{a\cdot b}{||a||||b||} = \frac{(0,4\sqrt[]{3},4)\cdot (0,0,1)}{8}= \frac{1}{2}$

Therefore, over the circle, $\phi = \frac{\pi}{3}$ . Note that rho varies from the plane z=4, up to the sphere, where rho is 8. Since $z = \rho \cos \phi$ , then over the plane we have that $\rho = \frac{4}{\cos \phi}$ Then, the following is the desired integral

$\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{3}}\int_{\frac{4}{\cos \phi}}^{8}\rho^2 \sin \phi d\rho d\phi d\theta$ where the new factor is the jacobian for the spherical coordinates.

d ) Let us use the integral in cylindrical coordinates

$\int_{0}^{2\pi}\int_{0}^{4\sqrt[]{3}} \int_{4}^{\sqrt[]{64-r^2}} rdz dr d\theta=\int_{0}^{2\pi}\int_{0}^{4\sqrt[]{3}} r (\sqrt[]{64-r^2}-4) dr d\theta=\int_{0}^{2\pi} d \theta \cdot \int_{0}^{4\sqrt[]{3}}r (\sqrt[]{64-r^2}-4)dr= 2\pi \cdot (-2\left.r^{2}\right|_0^{4\sqrt[]{3}})\int_{0}^{4\sqrt[]{3}}r \sqrt[]{64-r^2} dr$

Note that we can split the integral since the inner part does not depend on theta on any way. If we use the substitution $u = 64-r^2$ then $\frac{-du}{2} = r dr$ , then

$=-2\pi \cdot \left.(\frac{1}{3}(64-r^2)^{\frac{3}{2}}+2r^{2})\right|_0^{4\sqrt[]{3}}=\frac{320\pi}{3}$

3 0

2 years ago

A has the coordinates (–4, 3) and B has the coordinates (4, 4). If DO,1/2(x, y) is a dilation of △ABC, what is true about the im

fomenos

A dilation is a transformation $D_{o,k}(x,y)$ , with center O and a scale factor of k that is not zero, that maps O to itself and any other point P to P'. The center O is a fixed point, P' is the image of P, points O, P and P' are on the same line.

In a dilation of $D_{o, \frac{1}{2} }(x,y)$ , the scale factor, $\frac{1}{2}$ is mapping the original figure to the image in such a way that the distances from O to the vertices of the image are half the distances from O to the original figure. Also the size of the image is half the size of the original figure.

Therefore, <span>If $D_{o, \frac{1}{2} }(x,y)$ is a dilation of △ABC, the truth about the image △A'B'C'</span> are:

<span>AB is parallel to A'B'.

$D_{o, \frac{1}{2} }(x,y)=( \frac{1}{2}x, \frac{1}{2}y)$

The distance from A' to the origin is half the distance from A to the origin.</span>

3 0

2 years ago

Read 2 more answers

As a promotion, the first 50 customers who entered a certain store at a mall were asked to choose from one of two discounts. The

qaws [65]

Answer:

A) Yes, because P (F∩S) = 0

Step-by-step explanation:

Hello!

50 customers of a store were asked to choose between two discounts:

Discount 1: 20% off all purchases for the day.

Discount 2: 10% off all purchases for the week.

28 choose discount 1

22 choose discount 2

F: the selected person choose discount 1.

S: the selected person choose discount 2.

Two events are mutually exclusive when the occurrence of one of them prevents the other from occurring in one repetition of the trial and the intersection between these two events is void with zero probability of happening.

In this case, since the customers were asked to choose one out of the two events, if the customer chooses the first one, then he couldn't have chosen the second one and vice-versa. Then the intersection between these two events has zero probability, symbolically:

P(F∩S)=0

I hope it helps!

8 0

2 years ago

For a population with an unknown distribution, the form of the sampling distribution of the sample mean is _____.a. exactly norm

maxonik [38]

Answer:

Approximately normal for large sample sizes

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question:

The distribution is unknown, so the sampling distribution will only be approximately normal when n is at least 30.

So the correct answer should be:

Approximately normal for large sample sizes

7 0

2 years ago

The demand function for a type of portable radio is given by the model d=85−6x2, where d is measured in dollars and x is measure

Lostsunrise [7]

Answer:

Part A : The required Profit equation : $\text{Profit}= 60x-6x^3$

Part B : 0.303 million radios the profit became 18 million.

Part C : The graph is attached below.

Step-by-step explanation:

Given : The demand function for a type of portable radio is given by the model $d=85-6x^2$ , where d is measured in dollars and x is measured in millions of units. The production cost is $25.00 per radio.

Note: Given that profit = revenue - costs

Revenue is demand times number sold.

Costs are productions costs per number sold.

To find :

Part A: Write an equation giving profit as a function of x million radios sold.

Solution : Let x be the number of radios.

Revenue is demand times number sold

$\text{Revenue}= x\times (85-6x^2)$

$\text{Revenue}= 85x-6x^3$

Let the cost of x number of radios is 25x.

Profit = Revenue - Costs

$\text{Profit}= 85x-6x^3-25x$

$P(x)= 60x-6x^3$

The required Profit equation : $P(x)= 60x-6x^3$

Part B : The company currently produces 3 million radios and makes a profit of $18,000,000, but would like to scale back production. What lesser number of radios could the company produce to yield the same profit?

Solution : It is shown graphically which is attached in part c

When we scale back production we get that,

When we sold 0.303 million radios the profit became 18 million.

Part C : Give a graph for the profit and label the parts of the graph.

The graph is attached below which shows the profitable radios points.

3 0

2 years ago