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3 years ago

15

20 POINTS!!!! Which best describes the relationship between the successive terns in the sequence shown ? The common difference i

s -10. The common difference is 10. The common ratio is -9. The common ratio is 9.

2 answers:

olga2289 [7]3 years ago

4 0

The common difference is -10

lara31 [8.8K]3 years ago

3 0

Answer:

B

Step-by-step explanation:

9-(-1)=10

(-1)-(-11)=10

etc...

The common difference(subtraction) between the numbers is 10.

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75 POINT HURRY!! IF YOUR RIGHT YOU GET BRAINLEST

Svetllana [295]

since it is the distance up to 3 hours

35 *3 = 105

but the student is counting by 10's so it should be 110

8 0

3 years ago

Read 2 more answers

Can someone explain how to get the right answers thx

Free_Kalibri [48]

Idk exactly what those angles are called. I think supplementary interior. Which means they add up to 180°
So, (4x-2)+(6x-8)=180
Simplify. 10x-10=180
Now we need to get x by itself so add 10 to both sides to get 10x=190
Then divide by 10 to get x by itself x=190/10
So x=19

Hope this helped!!!

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3 years ago

What is the percent of 120 minus 94

lyudmila [28]

The correct answer is 26%

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3 years ago

Vonna scored 35 goals at soccer practice. Her ratio of goals to misses was 5:8. How many times did she miss?

Digiron [165]

<u>Answer:</u> She misses 56 times at a soccer practice.

<u>Step-by-step explanation:</u>

Let the goals she misses be 'x'.

According to the question:

$5:8::35:x$

These two ratios follow direct relationship, so:

$\frac{5}{8}=\frac{35}{x}\\\\5x=35\times 8\\\\x=\frac{280}{5}\\\\x=56$

Hence, Vonna misses 56 times at a soccer practice.

5 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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2 years ago