3 years ago

A rectangular pool is 30 1/3 feet long and 12 1/2 feet wide. What is the area of the pool?

Mathematics

1 answer:

butalik [34]3 years ago

4 0

About 379.1666666666666
The 6 is repeating. Or about 379 if your rounding.

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Divide the number 20 into two parts (not necessarily integers) in a way that makes the product of one part with the square of th

Ilya [14]

Answer:

13 1/3 and 6 2/3.

Step-by-step explanation:

Let the 2 parts be x and (20 - x).

So x^2(20 - x) must be a maximum.

y = x^2(20 - x)

y = 20x^2 - x^3

Finding the derivative):

y' = 40x - 3x^2 = 0 for maxm/minm value.

x( 40 - 3x) = 0

40 - 3x = 0

3x = 40

x = 40/3.

This is a maximum because the second derivative y" = 40 - 6x = 40 - 6(40/3)) is negative.

So the 2 numbers are 13 1/3 and 6 2/3.

3 0

2 years ago

1 pt) find the average rate of change of the function over the given intervals. h(t)=\cot (t);

blondinia [14]

$\text{Consider the function}\\
\\
h(t)=\cot(t) \text{ on the interval }\left [ \frac{\pi}{4}, \ \frac{3\pi}{4} \right ]\\
\\
\text{we know that the average rate of change of a funtion f(x) over the }\\
\text{interval [a, b] is given by}\\
\\
f_{avg}=\frac{f(b)-f(a)}{b-a}\\
\\
\text{so using this, the average rate of change of the given function is}$

$h_{avg}=\frac{h\left ( \frac{3\pi}{4} \right )-h\left ( \frac{\pi}{4} \right )}{\frac{3\pi}{4}-\frac{\pi}{4}}\\
\\
=\frac{\cot \left ( \frac{3\pi}{4} \right )-\cot \left ( \frac{\pi}{4} \right )}{\frac{3\pi-\pi}{4}}\\
\\
=\frac{(-1)-(1)}{\frac{2\pi}{4}}\\
\\
=\frac{-2}{\frac{\pi}{2}}\\
\\
\text{Average rate of change of function}=\frac{-4}{\pi}$