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3 years ago

14

the figure on the left represents a scale drawing 5yd. of the figure on the right 2 in. what is the scale?

1 answer:

Nikolay [14]3 years ago

6 0

Answer:

90 : 1

Step-by-step explanation:

1 yard = 36 inches

5 yards = 36×5 = 180 inches

180 : 2

90 : 1

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I think of a number. I multiply it by 7. I add 12. The result is 40. What is the number I am thinking of?

julia-pushkina [17]

The number that you are thinking of is 4. To get this answer I did the opposite of how you received your answer.
40 - 12 = 28
28 ÷ 4 = 7
The number that you are thinking of is 4.

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2 years ago

Linda is sending out cards.If envelopes come in boxes of 25 and stamps come in packs of 10,what is the least number of stamps an

chubhunter [2.5K]

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3 years ago

The boundary of a lamina consists of the semicircles y = 1 − x2 and y = 16 − x2 together with the portions of the x-axis that jo

oksano4ka [1.4K]

Answer:

Required center of mass $(\bar{x},\bar{y})=(\frac{2}{\pi},0)$

Step-by-step explanation:

Given semcircles are,

$y=\sqrt{1-x^2}, y=\sqrt{16-x^2}$ whose radious are 1 and 4 respectively.

To find center of mass, $(\bar{x},\bar{y})$ , let density at any point is $\rho$ and distance from the origin is r be such that,

$\rho=\frac{k}{r}$ where k is a constant.

Mass of the lamina=m= $\int\int_{D}\rho dA$ where A is the total region and D is curves.

then,

$m=\int\int_{D}\rho dA=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}rdrd\theta=k\int_{}^{}(4-1)d\theta=3\pi k$

Now, x-coordinate of center of mass is $\bar{y}=\frac{M_x}{m}$ . in polar coordinate $y=r\sin\theta$

$\therefore M_x=\int_{0}^{\pi}\int_{1}^{4}x\rho(x,y)dA$

$=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}(r)\sin\theta)rdrd\theta$

$=k\int_{0}^{\pi}\int_{1}^{4}r\sin\thetadrd\theta$

$=3k\int_{0}^{\pi}\sin\theta d\theta$

$=3k\big[-\cos\theta\big]_{0}^{\pi}$

$=3k\big[-\cos\pi+\cos 0\big]$

$=6k$

Then, $\bar{y}=\frac{M_x}{m}=\frac{2}{\pi}$

y-coordinate of center of mass is $\bar{x}=\frac{M_y}{m}$ . in polar coordinate $x=r\cos\theta$

$\therefore M_y=\int_{0}^{\pi}\int_{1}^{4}x\rho(x,y)dA$

$=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}(r)\cos\theta)rdrd\theta$

$=k\int_{0}^{\pi}\int_{1}^{4}r\cos\theta drd\theta$

$=3k\int_{0}^{\pi}\cos\theta d\theta$

$=3k\big[\sin\theta\big]_{0}^{\pi}$

$=3k\big[\sin\pi-\sin 0\big]$

$=0$

Then, $\bar{x}=\frac{M_y}{m}=0$

Hence center of mass $(\bar{x},\bar{y})=(\frac{2}{\pi},0)$

3 0

3 years ago

What’s the answer to this ?

Juli2301 [7.4K]

Answer:

the second one because 3 divided by 1.5 is 2 then it's right

7 0

3 years ago

Find the median of the data in the box plot below.

Dmitry_Shevchenko [17]

Median is just the middle number. If you put all the numbers in order from least to greatest:
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Then you can find the one in the middle. In this case it’s 7.

3 0

2 years ago

Read 2 more answers