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2 years ago

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I need help, you’ll receive brainlest!

1 answer:

OleMash [197]2 years ago

5 0

Answer:

D)

Step-by-step explanation:

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the third side of an isosceles triangle is 8 inches shorter than the other two sides the perimeter of the triangle is 70 inches

dezoksy [38]

Isosceles triangle has 2 sides that are the same length. So what you have is three sides that measure: x, x, and (x-8). 70 = x + x + (x-8).
70 = x + x + x - 8
78 = 3x
x = 26.
and 26 - 8 = 18.
So two of your sides are 26 inches and the last side is 18 inches.

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3 years ago

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Evaluate 4b – a when a = 12 and b=7.<br> 4b - a= 0

klio [65]

Answer: 4b - a = 15

Step-by-step explanation: if you replace the variables with their equivalent counterparts you get the equation 4(7) - 12 = __ when you do the math make sure to do the parentheses first then subtract, just like in elementary school remember pemdas parentheses, exponents, multiplication, division, addition, subtraction. I hope this helps.

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2 years ago

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Explain the different ways you can name an angle

nydimaria [60]

Answer:

When you say that, I think directly to Triangles. SO i am going to state that.

There is acute: Less than 90 degrees

There is Obtuse: more than 90 degrees

There is Right: 90 degrees

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3 years ago

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The sum of two integers is 21. Fifteen less than double the square of the smaller integer is equal to the larger

Salsk061 [2.6K]

Answer:

2x^2+x-36=0

(4,0)

2*4^2+4-15=21

Step-by-step explanation:

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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