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2 years ago

5

The playing surfaces of two foosball tables are similar. The ratio of the corresponding side lengths is 10 : 7. What is the rati

o of the areas ?

1 answer:

sweet [91]2 years ago

7 0

1 can be 10*7 the other 7*10

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Can anyone help with 19-22 with work shown I’m really confused

yanalaym [24]

Answer:

19. C

20. D

21. B

22. A

Step-by-step explanation:

If an angle is subtended from an arc and it goes to the opposite end of the circle, the angle is HALF of the measure of the arc. If it goes to the center, it is SAME measure as arc.

19.

Arc BC is DOUBLE of Angle BAC, so

78 * 2 = 156

Answer is C

20.

Arc BEC is DOUBLE of Angle BAC, so

114 * 2 = 228

Answer is D

21.

Angle ABC is HALF of Arc AC, so

Angle ABC = 110/2 = 55

Answer is B

22.

Angle ABC is HALF of Arc AC, so

Angle ABC = 218/2 = 109

Answer is A

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Cfrac%7B%281-x%5E%7B2%7D%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%20-x%5E%7B2%7D%20%281-x%5E%7B2%7D%

3241004551 [841]

Step-by-step explanation:

Hi, Hope this will help:)

6 0

3 years ago

A human bone is made up of 14 water and 920 minerals, and the rest is living tissue. What decimal represents the amount of a hum

Cerrena [4.2K]

D because multiple then divided

3 0

2 years ago

Ella has a part-time job selling cupcakes at a cart in the park. She is paid $25 for a shift plus a fee of $0.25 for each cupcak

mojhsa [17]

9514 1404 393

Answer:

A) t = 0.25c +25

B) c = 52

C) 3 weekday shifts

Step-by-step explanation:

A. The amount Ella earns is $0.25 per cupcake, so for c cupcakes, that is ...

0.25c

In addition, she is paid $25, so her total amount she is paid is ...

t = 0.25c +25

__

B. In order to earn $38, the number of cupcakes sold must be ...

38 = 0.25c +25

13 = 0.25c . . . . . . subtract 25

52 = c . . . . . . . . . multiply by 4

Ella sold 52 cupcakes to earn $38.

__

C. On one of the average weekdays, Ella could be expected to earn ...

t = 0.25(36) +25 = 34

Then for 3 weekday shifts, her earnings would be 3($34) = $102. (It would take Ella 3 weeks to earn $300.)

On one of the average weekend shifts, Ella could be expected to earn ...

t = 0.25(62) +25 = 40.50

Then for 2 weekend shifts, her earnings would be 2($40.50) = $81. (It would take Ella 4 weekends to earn $300.)

Ella would earn $300 more quickly by working weekday shifts.

8 0

3 years ago

99 POINT QUESTION, PLUS BRAINLIEST!!!

VladimirAG [237]

First, we have to convert our function (of x) into a function of y (we revolve the curve around the y-axis). So:

$y=100-x^2\\\\x^2=100-y\qquad\bold{(1)}\\\\\boxed{x=\sqrt{100-y}}\qquad\bold{(2)} \\\\\\0\leq x\leq10\\\\y=100-0^2=100\qquad\wedge\qquad y=100-10^2=100-100=0\\\\\boxed{0\leq y\leq100}$

And the derivative of x:

$x'=\left(\sqrt{100-y}\right)'=\Big((100-y)^\frac{1}{2}\Big)'=\dfrac{1}{2}(100-y)^{-\frac{1}{2}}\cdot(100-y)'=\\\\\\=\dfrac{1}{2\sqrt{100-y}}\cdot(-1)=\boxed{-\dfrac{1}{2\sqrt{100-y}}}\qquad\bold{(3)}$

Now, we can calculate the area of the surface:

$A=2\pi\int\limits_0^{100}\sqrt{100-y}\sqrt{1+\left(-\dfrac{1}{2\sqrt{100-y}}\right)^2}\,\,dy=\\\\\\= 2\pi\int\limits_0^{100}\sqrt{100-y}\sqrt{1+\dfrac{1}{4(100-y)}}\,\,dy=(\star)$

We could calculate this integral (not very hard, but long), or use (1), (2) and (3) to get:

$(\star)=2\pi\int\limits_0^{100}1\cdot\sqrt{100-y}\sqrt{1+\dfrac{1}{4(100-y)}}\,\,dy=\left|\begin{array}{c}1=\dfrac{-2\sqrt{100-y}}{-2\sqrt{100-y}}\end{array}\right|= \\\\\\= 2\pi\int\limits_0^{100}\dfrac{-2\sqrt{100-y}}{-2\sqrt{100-y}}\cdot\sqrt{100-y}\cdot\sqrt{1+\dfrac{1}{4(100-y)}}\,\,dy=\\\\\\ 2\pi\int\limits_0^{100}-2\sqrt{100-y}\cdot\sqrt{100-y}\cdot\sqrt{1+\dfrac{1}{4(100-y)}}\cdot\dfrac{dy}{-2\sqrt{100-y}}=\\\\\\$

$=2\pi\int\limits_0^{100}-2\big(100-y\big)\cdot\sqrt{1+\dfrac{1}{4(100-y)}}\cdot\left(-\dfrac{1}{2\sqrt{100-y}}\, dy\right)\stackrel{\bold{(1)}\bold{(2)}\bold{(3)}}{=}\\\\\\= \left|\begin{array}{c}x=\sqrt{100-y}\\\\x^2=100-y\\\\dx=-\dfrac{1}{2\sqrt{100-y}}\, \,dy\\\\a=0\implies a'=\sqrt{100-0}=10\\\\b=100\implies b'=\sqrt{100-100}=0\end{array}\right|=\\\\\\= 2\pi\int\limits_{10}^0-2x^2\cdot\sqrt{1+\dfrac{1}{4x^2}}\,\,dx=(\text{swap limits})=\\\\\\$

$=2\pi\int\limits_0^{10}2x^2\cdot\sqrt{1+\dfrac{1}{4x^2}}\,\,dx= 4\pi\int\limits_0^{10}\sqrt{x^4}\cdot\sqrt{1+\dfrac{1}{4x^2}}\,\,dx=\\\\\\= 4\pi\int\limits_0^{10}\sqrt{x^4+\dfrac{x^4}{4x^2}}\,\,dx= 4\pi\int\limits_0^{10}\sqrt{x^4+\dfrac{x^2}{4}}\,\,dx=\\\\\\= 4\pi\int\limits_0^{10}\sqrt{\dfrac{x^2}{4}\left(4x^2+1\right)}\,\,dx= 4\pi\int\limits_0^{10}\dfrac{x}{2}\sqrt{4x^2+1}\,\,dx=\\\\\\=\boxed{2\pi\int\limits_0^{10}x\sqrt{4x^2+1}\,dx}$

Calculate indefinite integral:

$\int x\sqrt{4x^2+1}\,dx=\int\sqrt{4x^2+1}\cdot x\,dx=\left|\begin{array}{c}t=4x^2+1\\\\dt=8x\,dx\\\\\dfrac{dt}{8}=x\,dx\end{array}\right|=\int\sqrt{t}\cdot\dfrac{dt}{8}=\\\\\\=\dfrac{1}{8}\int t^\frac{1}{2}\,dt=\dfrac{1}{8}\cdot\dfrac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}=\dfrac{1}{8}\cdot\dfrac{t^\frac{3}{2}}{\frac{3}{2}}=\dfrac{2}{8\cdot3}\cdot t^\frac{3}{2}=\boxed{\dfrac{1}{12}\left(4x^2+1\right)^\frac{3}{2}}$

And the area:

$A=2\pi\int\limits_0^{10}x\sqrt{4x^2+1}\,dx=2\pi\cdot\dfrac{1}{12}\bigg[\left(4x^2+1\right)^\frac{3}{2}\bigg]_0^{10}=\\\\\\= \dfrac{\pi}{6}\left[\big(4\cdot10^2+1\big)^\frac{3}{2}-\big(4\cdot0^2+1\big)^\frac{3}{2}\right]=\dfrac{\pi}{6}\Big(\big401^\frac{3}{2}-1^\frac{3}{2}\Big)=\boxed{\dfrac{401^\frac{3}{2}-1}{6}\pi}$

Answer D.

6 0

3 years ago

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