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grandymaker [24]
3 years ago
6

Julio says, "If you subtract 17 from my number and multiply the difference by - 3, the result is - 36." What is Julio's number?

​
Mathematics
2 answers:
adelina 88 [10]3 years ago
7 0
It would be 29, to solve this, you just have to reverse everything that was said, so there you would divide -36 by 3 and add 17
ValentinkaMS [17]3 years ago
5 0

Answer:

Julio’s number is 29

Step-by-step explanation:

(x - 17) · -3 = -36

x - 17 = 12

x = 29

<u>Check</u>:

29 - 17 = 12 · -3 = -36

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Alina [70]
The answer is -24+8= -16
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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
3 years ago
Help please I will give you the brainliest
ioda

Answer:2

Step-by-step explanation:

4 0
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Brianna has six bags of soil filling one flower pot requires 1/3 of a bag of soil how many flower pots can be filled with the si
dimaraw [331]

Answer:

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Step-by-step explanation:

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Keep Change Flip

\frac{6}{1} × \frac{3}{1} = \frac{18}{1}

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$22.00+$14.00= <br> answer this
Zigmanuir [339]

Answer:

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Step-by-step explanation:

22+14=36

6 0
3 years ago
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