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3 years ago

15

-12/5 divided by -1/5

1 answer:

Sindrei [870]3 years ago

6 0

The answer is 12!
You’re welcomeeee

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Find the equation of the line using the given information.

Dominik [7]

Answer:

x 1 = -3

x2 = -3

y1 = 0

y2 = 5

The slope of that equation is infinite and the equation is a perfectly vertical line at x = -3.

So, the equation is x = -3.

Step-by-step explanation:

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3 years ago

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How do u estimate 5004

Brrunno [24]

Answer:

$5004 \approx5000$

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3 years ago

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How long does it take a 2700 dollar investment compounded monthly at 9% to make 500 dollars

valina [46]

Answer:

23 months

Also please note that it's everytime ^t/12 not (^t)/12

Step-by-step explanation:

Principal (P) = $2700

rate = 9%

t = ?

Amount = $P(1+\frac{r}{100} )^(t/12)$

Amount = 2700 +500 = $3200

So substituting all our values we get :

$3200 = 2700(1+\frac{9}{100})^(t/12)\\ 3200=2700(1+0.09)^(t/12)\\3200 = 2700(1.09)^(t/12)\\\frac{3200}{2700} =1.09^(t/12)$

So we use log to solve for t :

$t/12=log_{1.09} (\frac{3200}{2700})\\$

Solving this using a calculator we get:

1.9714966193*12

=23 months

7 0

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 $(\bar{x},\bar{y})=(\frac{2}{\pi},0)$

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3 years ago

How to write two and a half million

lakkis [162]

Answer:

2,500,000

Hope this helps! ;)

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3 years ago

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