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2 years ago

9

Evaluate the expression a=7 and x = -6

1 answer:

Sergeeva-Olga [200]2 years ago

8 0

Answer:

i need more information to solve this

Step-by-step explanation:

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You are building a tower that is 4 ft. tall. Which would your need more of to build your 4 ft. tall tower?

laiz [17]

Answer:

pennies

Step-by-step explanation:

because they are thinner than notebooks and you will need more of them

3 0

2 years ago

300 is what percent less than 599

padilas [110]

Answer:

59.8%

Step-by-step explanation:

599 - 300 = 299. Now we need to find the percentage 299 is of 500.

299 / 500 = 0.598

0.598 as a percent is 59.8%.

Answer: 59.8%

Hope this helps!

7 0

3 years ago

. Simplify the expression: 4<br> 2 + 8 ÷ 2.

zhannawk [14.2K]

Answer:

42+4

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

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

Tasha purchased a 40.00$ shirt on sale for 30.00$ how much of a discount did tasha receive.

expeople1 [14]

Answer:

The shirt is actually 25% off.

Step-by-step explanation:

While the shirt obviously lost a quarter of its price, you could also work the equation:

10 / 40 = 25

6 0

3 years ago