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3 years ago

12

Please help me please

1 answer:

egoroff_w [7]3 years ago

8 0

Answer:

y = 4 (Answer C)

Step-by-step explanation:

We want to find side y. The sine function relates the 30 degree angle, the hypotenuse 8 and the opposite side y:

y

sin 30 degrees = 1/2 = -----------

8

so y = 4 (Answer C)

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Find the area of the parking space if the height is 18 feet and the base is 9 1/4 feet

Molodets [167]

Area= base x height
So, the base is 9 1/4 and the height is 18 feet
so, the Area= 9 1/4 x 18

9 1/4 x 18
= 166.5

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3 years ago

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2 years ago

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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

find the slope of the line containing the points (-1,6) and (3,5) put final.answer in slope-intercept form

Ugo [173]

Given the points (-1,6) and (3,5).

The formula to find the slope is

$m=\frac{y_2-y_1_{}}{x_2-x_1}$

Take

$x_1=-1,y_1=6,x_2=3,y_2=5$

Plug the values into the formula and find the slope.

$\begin{gathered} m=\frac{5-6}{3-(-1)} \\ =\frac{-1}{4} \end{gathered}$

The slope-intercept form is y = mx+b.

Plug the value of m.

$y=-\frac{1}{4}x+b$

ThusConsider the point (-1,6). Substitute -1 for x and for y into the equation.

$\begin{gathered} 6=-\frac{1}{4}(-1)+b \\ =\frac{1}{4}+b \\ b=6-\frac{1}{4} \\ =\frac{23}{4} \end{gathered}$

Thus, the equation of the line in slope intercept form is

$y=-\frac{1}{4}x+\frac{23}{4}$

3 0

7 months ago