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4 years ago

I need help with #20

Mathematics

3 answers:

lora16 [44]4 years ago

8 0

Hey there!

Formula given :

A = πr²

Solve for r

A / π = r²

r = √( A / π )

Hope it helps!

g100num [7]4 years ago

5 0

Hope this helps.
i have solved the equation to help you get a better of how to demonstrate this in the future.

Darina [25.2K]4 years ago

3 0

A=pi r^2
sqrt(A)=pi r
sqrt(A)/pi=r

Answer:sqrt(A)/pi=r

(sqrt(A) means the square root of A)

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on a tour of an old gold mine, you find a gold nugget containing .82 ounces of gold. gold is worth $1566.80 per ounce. how much

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1566.80 * 0.82 = 1331.77

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4 years ago

Read 2 more answers

Two functions are given below. How does the graph of

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I think it’s C. I plugged it on to a graph and shows that graph a is steeper than graph b. red is graph a. blue is graph b.

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3 years ago

A baker bakes 50 muffins. 1/5 of the muffins are chocolate chip. 1/2 of the muffins are blueberry. The rest are cinnamon. How ma

vitfil [10]

Answer:

15 muffins are cinnamon.

Step-by-step explanation:

Given that:

Number of muffins baked = 50 muffins

Chocolate chip muffins = 1/5 of 50 = $\frac{1}{5}*50$

Chocolate chip muffins = 10 muffins

Blueberry muffins = 1/2 of 50 = $\frac{1}{2}*50$

Blueberry muffins = 25

Cinnamon muffins = Muffins baked - chocolate chip muffins - blueberry muffins

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Cinnamon muffins = 50 - 35 = 15

Hence,

15 muffins are cinnamon.

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

$\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}$

7 0

2 years ago

Events A and B have probabilities such that P(A) = 0.3, P(B) = 0.25, P(M ∪ N) = 0.425, and P(M ∩ N) = 0.075. Are events A and ev

nekit [7.7K]

Answer:

(C)Yes, because P(M) ∙ P(N) = P(M ∩ N)

Step-by-step explanation:

Two events A and B are independent if P(A)P(B)=P(A ∩ B)

Given events A and B such that:

P(A) = 0.3, P(B) = 0.25, P(A ∪ B) = 0.425, and P(A ∩ B) = 0.075

P(A)P(B)=0.3 X 0.25 =0.075
P(A ∩ B) = 0.075

Since the two expression above gives the same answer, they are independent.

<u>The correct option is C.</u>

7 0

4 years ago