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Grace [21]
3 years ago
11

Nth term of term 5 10 15 20 25

Mathematics
1 answer:
nekit [7.7K]3 years ago
7 0

Answer:

  5n

Step-by-step explanation:

The sequence is a list of multiples of 5, starting with 5×1. The n-th multiple is 5n.

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Is 5 1/ 5 to the third power equal to 5 (5 to the third power) explain
swat32

Answer:

Step-by-step explanation:

If I read this correctly...

5 1/5^3= 5.20^3=140.608

5(5)^3=625

If this isn't the correct problems, sorry. it was kind of hard to decipher the last one.

4 0
2 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
2 years ago
I need help with Question 12
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Keep in mind both the distance and time of the scenarios. The faster and more distance travelled = a higher gradient for the line. If he slows, the gradient decreases. If he stops, the line is flat but time still goes on.
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Please look at photo and ignore already selected answer
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Answer:

Question (A) firgue b is pallorgram

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Step-by-step explanation:

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Use PEMDAS to solve this problem <br><br> 6÷2(1+2)=?
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The answer would be 8. hope this helps
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2 years ago
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