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lorasvet [3.4K]

3 years ago

11

2 Points Factor the trinomial below. 9x2 + 12x+ 4

1 answer:

denis-greek [22]3 years ago

4 0

Answer:

(3x-2)^2

Step-by-step explanation:

9x^2+6x+6x+4 now factor the first and second pair of terms...

3x(3x+2)+2(3x+2) which is equal to:

(3x+2)(3x+2) which is just:

(3x+2)^2

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Area of a pentagon with a radius of 8m

san4es73 [151]

Area = [radius^2 * 5 * sine (360/5)] / 2
area = [64 * 5 * 0.95106] / 2
area = <span> <span> <span> 152.1696 </span> </span> </span>

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3 years ago

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Sergeeva-Olga [200]

Answer:

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Step-by-step explanation:

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3 years ago

A sphere has a diameter of 10 in. What is the volume of the sphere?

givi [52]

The volume of sphere is $\frac{500}{3}\pi$ inches cubed.

Step-by-step explanation:

Given,

Diameter of sphere = 10 inches

Radius of sphere = $\frac{Diameter}{2}=\frac{10}{2}$

Radius of sphere = 5 inches

Volume of sphere = $\frac{4}{3}\pi r^3$

V = $\frac{4}{3}\pi (5)^3$

$V = \frac{4}{3}\pi * 125\\\\V = \frac{500}{3}\pi$

The volume of sphere is $\frac{500}{3}\pi$ inches cubed.

Keywords: volume, division

Learn more about division at:

#LearnwithBrainly

5 0

3 years ago

Read 2 more answers

How many cars did he put in each box

Step2247 [10]

The answer is 8 cars.

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3 years ago

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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