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3 years ago

Help please!!!!!!!!!!!

Mathematics

1 answer:

Lisa [10]3 years ago

8 0

Answer:

x = 7

Step-by-step explanation:

If ∠ABD has a measure of 3x and ∠ABD is half of ∠ABC, then angle ABC has a measure of 6x. If ∠ABC measures 42°, then 6x = 42.

Divide both sides by 6.

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Step-by-step explanation:

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Subtract and simplify<br> (y^2 – 3y – 5) - (-y^2 – 7y+4)

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3 years ago

Give the domain and range.

MA_775_DIABLO [31]

Answer:

Domain: {-3, 0 3}

Range: {0, 3}

Step-by-step explanation:

Recall that the Domain of a relationship associated with coordinate pairs of the form (x,y), is the set of all "x-values" given in the relationship.

The Range is the set of all "y-values" given in the relationship.

In our case, we have three coordinate points defining the relationship. These points have the following (x,y) coordinates (see attached image to see which coordinates are associated with each point):

(-3, 3), (0, 0), and (3, 3)

Therefore, the x values associated with these points are: -3, 0 and 3

therefore the Domain of the relationship is the set: { -3, 0, 3}

On the other hand, he y values associated with these points are: 3, 0, and again 3. So in fact there are just to distinct y-values: 0 and 3.

therefore the Range of the relationship is he set: {0, 3}

which corresponds to answer B.

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

3 0

3 years ago