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Step2247 [10]
3 years ago
9

9a- 24 >48 what is the solution

Mathematics
2 answers:
Artyom0805 [142]3 years ago
8 0

Answer:

The solution to the given inequality is a > 8.

Step-by-step explanation:

For this problem, we have to solve for the inequality.

9a - 24 > 48

add 24 on both sides of the inequality.

9a > 72

Divide 9 on both sides of the inequality.

a > 8

So, your solution would be a > 8.

chubhunter [2.5K]3 years ago
5 0

Answer:

a > 8

Step-by-step explanation:

To solve this inequality, we have to get the variable, <em>a, </em>by itself on one side of the equation

9a -24 > 48

24 is being subtracted from 9a. The inverse of subtraction is addition. Add 24 to both sides of the equation.

9a-24+24 > 48+24

9a > 48 +24

9a > 72

a is being multiplied by 9. The inverse of multiplication is division. Divide both sides of the equation by 9.

9a/9 > 72/9

a > 72/9

a > 8

The solution to this inequality is a > 8

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Share 747 in the ratio 2:7 between Tom and Ben
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Answer:

tom = 166, ben = 581

Step-by-step explanation:

2 + 7 = 9

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Calculus Problem
Roman55 [17]

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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
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