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2 years ago

This is past due I need help will give brainlist

Mathematics

2 answers:

Licemer1 [7]2 years ago

6 0

Answer:

Step-by-step explanation:

Flauer [41]2 years ago

5 0

Your answer is A hope this helps! Have a great day!

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Calculus Problem

Roman55 [17]

The two parabolas intersect for

$8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2$

and so the base of each solid is the set

$B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}$

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, $|x^2-(8-x^2)| = 2|x^2-4|$ . But since -2 ≤ x ≤ 2, this reduces to $2(x^2-4)$ .

a. Square cross sections will contribute a volume of

$\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x$

where ∆x is the thickness of the section. Then the volume would be

$\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}$

where we take advantage of symmetry in the first line.

b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

$\dfrac\pi8 \left(2(x^2-4)\right)^2 \, \Delta x = \dfrac\pi2 (x^2-4)^2 \, \Delta x$

We end up with the same integral as before except for the leading constant:

$\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx$

Using the result of part (a), the volume is

$\displaystyle \frac\pi8 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{256\pi}{15}}}$

c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

$\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x$

and using the result of part (a) again, the volume is

$\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}$

7 0

2 years ago

Three side lengths of a right triangle are given which side length should you substitute for the hypotenuse in Pythagorean theor

ankoles [38]

Answer:

The longest length which you are given.

Step-by-step explanation:

The hypotenuse is the longest side in a right angled triangle. The largest side in a right angled triangle is known as the hypotenuse.

8 0

2 years ago

How to solve y-b/x=m for y

Slav-nsk [51]

To solve for m we want to get m by itself on one side of the equals sign, with everything else on the other side. First we subtract b from both sides, to get it out of the right side of the equation:
y - b = mx + b - b
y - b = mx
Now we divide both sides by x, to get m completely by itself:
(y-b)/x = (mx)/x
(y-b)/x = m
So m is equal to (y-b)/x.

8 0

3 years ago

Read 2 more answers

What fraction is equivalent to 3/9

cluponka [151]

Answer:

There are infinity equivalent fractions to 3/9

Step-by-step explanation:

1 /3, 2 /6, 3/ 9, 4 /12, 5/ 15, 6 /18, 7/ 21, 8/ 24, 9 /27, 10/ 30, 11/ 33, 12/ 36, 13/ 39, 14 /42, 15 /45..........

7 0

2 years ago

Oil is pumped continuously from a well at a rate proportional to the amount of oil left in the well. Initially there were millio