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3 years ago

Answer questions below ASAP 5-1/2x=5/8x+2

Mathematics

1 answer:

masya89 [10]3 years ago

4 0

5 - (1/2)x = (5/8)x + 2
5-2 = (5/8)x + (1/2)x
3 = (9/8)x
(8*3)/9 = x
8/3 = x
x = 2/2/3

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Solve the following equation-<br> p-4=(-9)+p

Mashcka [7]

Simplifying p + -4 = (-9) + p
Reorder the terms: -4 + p = (-9) + p -4 + p = -9 + p
Add '-1p' to each side of the equation. -4 + p + -1p = -9 + p + -1p
Combine like terms: p + -1p = 0 -4 + 0 = -9 + p + -1p -4 = -9 + p + -1p
Combine like terms: p + -1p = 0 -4 = -9 + 0 -4 = -9
Solving -4 = -9

3 0

3 years ago

The diameter of the moon is 2,160 miles. A model has a scale of 1 in : 150 mi. What is the diameter of the model?

Charra [1.4K]

Answer:

The diameter of the model is 14.4 inches.

Step-by-step explanation:

The Diameter of the moon = 2,160 miles

The scale on the model represents 1 in = 150 miles

Let the model represents k inches in 2,160 miles.

So, by the Ratio of Proportionality:

$\frac{1}{150} = \frac{k}{2160}$

⇒ $k = \frac{2160}{150} \\\implies k = 14.4$

or, k = 14.4 inches

⇒On the scale 2160 miles is represented as 14.4 inches

Hence the diameter of the model is 14.4 inches.

3 0

3 years ago

Electric charge is distributed over the disk x2 + y2 ≤ 16 so that the charge density at (x, y) is rho(x, y) = 2x + 2y + 2x2 + 2y

professor190 [17]

Answer:

Required total charge is $256\pi$ coulombs per square meter.

Step-by-step explanation:

Given electric charge is dristributed over the disk,

$x^2=y^2\leq 16$ so that the charge density at (x,y) is,

$\rho (x,y)=2x+2y+2x^2+2y^2$

To find total charge on the disk let Q be the total charge and $x=r\cos\theta,y=r\sin\theta$ so that,

$Q={\int\int}_Q\rho(x,y) dA$ where A is the surface of disk.

$=\int_{0}^{2\pi}\int_{0}^{4}(2x+2y+2x^2+2y^2)dA$

$=\int_{0}^{2\pi}\int_{0}^{4}(2r\cos\theta+2r\sin\theta+2r^2 \cos^{2}\theta+2r^2\sin^2\theta)rdrd\theta$

$=2\int_{0}^{2\pi}\int_{0}^{4}r^2(\cos\theta+\sin\theta)drd\theta+2\int_{0}^{2\pi}\int_{0}^{4}r^3drd\theta$

$=\frac{2}{3}\int_{0}^{2\pi}(\sin\theta+\cos\theta)\Big[r^3\Big]_{0}^{4}d\theta+2\int_{0}^{2\pi}\Big[\frac{r^4}{4}\Big]d\theta$

$=\frac{128}{3}\int_{0}^{2\pi}(\sin\theta+\cos\theta)d\theta+128\int_{0}^{2\pi}d\theta$

$=\frac{128}{3}\Big[\sin\theta-\cos\theta\Big]_{0}^{2\pi}+128\times 2\pi$

$=\frac{128}{3}\Big[\sin 2\pi-\cos 2\pi-\sin 0+\cos 0\Big]+256\pi$

$=256\pi$

Hence total charge is $256\pi$ coulombs per square meter.

3 0

3 years ago

Two trucks are driving to the same place. The first truck starts 50 miles ahead of the second truck and travels at an average sp

Dominik [7]

Well obviously the two bottom graphs don't represent the two trucks meeting. So we can cross those out. Looking at the top two graphs we can tell that the line above is truck #2. In graph 1 it shows the second truck is slowing down which is wrong since that isn't told.

So the answer would be Graph #2

I hope this helps!

-Ayden

5 0

3 years ago

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Kendra walked diagonally across a rectangular field that is 45 m by 65 m. To the nearest tenth of a meter, how far did she walk?

Crazy boy [7]

Answer:

Step-by-step explanation:

4 0

3 years ago