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3 years ago

Please help me I’m about to graduate

Mathematics

1 answer:

vivado [14]3 years ago

4 0

Answer:

B & D

Step-by-step explanation:

If you do 30x10 it would equal 300

So that wouldn't be it. If you do 49x100 that will equal 4,900 which is correct. If you do 40x600 that would equal 24,000 which is also incorrect. The last one is correct because 400x80 is 32,000

(Your welcome!)

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PLS THIS IS MY LAST QUESTION

MrRa [10]

Answer: x = 5.625

Step-by-step explanation: The triangles are similar; the side lengths of the large triangle are proportional to the lengths of the small triangle. The corresponding parts are proportional.

Use proportion. The total length of side FV is 16.

6:16 = x:15 . or 6/16=x/15

cross multiply 15 × 6 = 90, then divide 90/16

90/16 = x

5.625 = x

5 0

3 years ago

8. Is line / parallel to line m? Explain. (1 point)

puteri [66]

Answer:

what lines..? post picture.

Step-by-step explanation:

7 0

3 years ago

Find the unknown measures. Round lengths to the nearest hundredth and angle measures to the nearest degree.

MatroZZZ [7]

The answer is AB= 4.62; mA= 22; mB=68
You could quickly find the answer by just using Pythagorean theorem to find AB.
4.3^2 +1.7^2 = AB^2

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

3 0

3 years ago

What is v29?

satela [25.4K]

Answer:

Please read explanations

Step-by-step explanation:

By direct calculations, we know that the nearest perfect square to 29 is 25

The next is 36;

Hence, since 29 is between 25 and 36; it is only proper that the square root of 29 is also between the square roots of both numbers

So, we can conclude that the square root is between 5 and 6

For actual values by calculating, we have this as 5.38 which to the nearest number is 5.4

so, we have the dot at the next 4 small lines after the number 5

7 0

3 years ago