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yawa3891 [41]

3 years ago

ss="latex-formula">

Mathematics

1 answer:

Georgia [21]3 years ago

5 0

Answer:

answer=cos2 I ---------1

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The volume of the box is modeled by the function V (x) = 5x²+ 5x. whats the volume when x=4

Reika [66]

Answer:

100

Step-by-step explanation:

V (x) = 5x²+ 5x

x=4

We plug in the x value:

V (x) = 5(4)²+ 5(4)

Now we sovle the equation:

5(4)² + 5(4)

5(16) + 5(4)

80 + 20

= 100

Hope that helps, and sorry if its incorrect

5 0

4 years ago

Factor 140c + 28 -14a to identify the equivalent expressions.

Vlada [557]

Step-by-step explanation:

140c+28-14a
14(10c+2-a)

hope it helps

<h2>stay safe healthy and happy...</h2>

8 0

3 years ago

A whole pie is cut into 8 equal slices. Three of the pie is left

bagirrra123 [75]

Answer:

Five is taken away

Step-by-step explanation:

8 0

3 years ago

Explain the difference between the translation of y=x² for the graphs of y=(x+3)² and y=x²+3.

LUCKY_DIMON [66]

Answer:

y=(x+3)^2 has a vertex at (-3,0), while y=x^2 + 3 has a vertex at (0,3)

Mark me brainliest hope this helps

Step-by-step explanation:

3 0

3 years ago

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

3 years ago