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2 years ago

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A teacher wanted to buy a chair, a bookshelf, two tables and a desk. She spent $900 for all five items and the chair and the des

k combined 70% of her total. If the bookshelf cost $50, how much did each of the tables cost?

1 answer:

mars1129 [50]2 years ago

8 0

Find the cost of the chair and desk by multiplying the total amount pent by the 70%

Chair and desk = 900 x 0.70 = $60

Subtract that cost from the total spent:

900 - 630 = $270 was spent on the other three items.

Subtract the cost of the bookshelf:

270 - 50 = 220

The two tables cost 220.

For the price of one table divide that cost by 2:

220/2 = 110

One table cost $110.

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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).

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

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

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d ) Let us use the integral in cylindrical coordinates

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