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Black_prince [1.1K]
3 years ago
5

No Links!! A tissue box has a height of 4 inches, a width of 8 inches and a length of 5.5 inches. What is the volume in inches c

ubed?
Mathematics
1 answer:
ANEK [815]3 years ago
7 0

Answer:

Step-by-step explanation: bey said yass

So it’s 89

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the sum of the first 6 terms of a geometric series is 15624 and the common ratio is 5. what is the first term of the series?
9966 [12]

Answer:

first term a=4

Step-by-step explanation:

Sum of the first 6 terms S6 = 15624

common ratio r= 5

n= 6

first term a=?

Formula: S6= a(r^n-1)/r-1

15624 = a( 5^6-1)/5-1

15624 = a(15625-1)/4

15624= a( 15624)÷4

15624= 15624a÷4

Cross multiply

15624×4 = 15624a

62496 = 15624a

divide both sides by 15624

62496÷15624 = 15624a÷15624

4 = a

a= 4

7 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}}

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19+19= x

So x, is 38. Just add 19 and 19

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