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2 years ago

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Point M lies on the line segment connecting L(-4, 2) and N (2,-4). If M partitions the segment in the ratio 1:2, what are the co

ordinates of M?

1 answer:

taurus [48]2 years ago

5 0

AWSER is C
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30000x^2 + 2x + 1200

jasenka [17]

Answer: 2(15000x2)+x+600)

Step-by-step explanation:

8 0

2 years ago

if an ice cream factory makes 350 quart of ice cream in 2 hours?how many quarts could be made in 12 hours? what was the rate in

UkoKoshka [18]

2,100 quarts could be made in 12 hours. The rate for that day was 175 quarts every hour. We get this by dividing 350 by 2.

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2 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

2 years ago

Confidence intervals for the means provide an estimate for where the true mean lies.

stepladder [879]

It is true that the confidence intervals for the mean provide an estimate for where the true mean lies.

In statistics, a confidence interval denotes the likelihood that a population parameter will fall between a set of values for a given proportion of the time. A confidence interval depicts the likelihood that a parameter will fall between two values near the mean. Confidence intervals quantify the degree of uncertainty or certainty in a sampling procedure.

The mean is a basic mathematical average of two or more values. There are two sorts of means that may be calculated: the arithmetic mean and the geometric mean. A mean tells you the average of a bunch of values, which helps you contextualize each data point.

To learn more about confidence interval, visit :

brainly.com/question/24131141

#SPJ4

3 0

11 months ago

Please help!!!!! I need help badly

Nata [24]

Answer:

3/4

Step-by-step explanation:

54/72 simplified is the same as 3/4

5 0

2 years ago