Ebase of a circular fence with radius m is given by , . the height of the fence at position is given by the function , so the height varies from m to m. suppose that l of paint covers . sketch the fence and determine how much paint you will need if you paint both sides of the fence.

The question is missing important equations. The complete question is:

Base of a circular fence with radius 10 m is given by x = 10 cos(t), y = 10 sin(t). the height of the fence at position (x, y) is given by the function h(x, y) = 3 + 0.03(x^2 − y^2), so the height varies from 0 m to 6 m. suppose that 1 L of paint covers 100 m^2. determine how much paint you will need if you paint both sides of the fence. (round your answer to two decimal places.)

0.38 liters of paint is needed to cover both sides of the fence.

Step-by-step explanation:

Given:

Radius of circular base is,

The height of the fence varies as:

------------1

Where, 'x' and 'y' varies with 't' as:

First, we need to write only as a function of , As the base of the fence is circular, we use the circle's equation:

Now, plugging the value of in equation (1), we get:

Now, we know that:

Therefore, the height can be expressed in terms of 't' as:

In order to find the area, we need to integrate equation (2).

The above area is the area of one side of the fence.

Therefore, the total area to be painted is 12π m².

Now, For 100 m² of area, volume of paint needed is 1 L.

So, for 1 m² of area, volume of paint needed is

Therefore, for 12π m² of area, volume of paint needed is:

Hence, 0.38 liters of paint is needed to cover both sides of the fence.

the mean and mad

step-by-step explanation:

your answer is a

step-by-step explanation:

4*4*4*4=256