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3 years ago

Someone please help me out

Mathematics

1 answer:

faust18 [17]3 years ago

7 0

Answer:

6*x^(2/5)*y^(1/5)

Step-by-step explanation:

The fifth root of x^2 is x^(2/5).

That of y is y^(1/5).

Then the given expression, rewritten with fractional exponents, is

6*x^(2/5)*y^(1/5)

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I need help please:)

UkoKoshka [18]

Find all the common factors ... answer is 2u^3v^2

4 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

What is the quotient to four dived by fifty

allsm [11]

0.08 if 4 divided by 50
12.5 if 50 divided by 4

8 0

3 years ago

Read 2 more answers

Which value of a in the exponential function below would cause the function to stretch?

mote1985 [20]

Answer:

a' > a

Step-by-step explanation:

$f(x)=ae^{bx}$

A larger value for a will stretch the exponential in the y direction. A larger value for b will stretch it in the x direction.

5 0

3 years ago

The mean salary of 5 employees is $33700. The median is $34600. The mode is $35600. If the median paid employee gets a $3500 rai

fgiga [73]

Step-by-step explanation:

Given:

Mean = 33700

Median = 34600

Mode = 35600

The mean is the average, the median is the middle number, and the mode is the most common number.

First, we need to find the new mean (average) if one of the employees gets a 3500 raise. The average is the total salary divided by number of employees:

(5 × 33700 + 3500) / 5 = 34400

The mode is the most common number in a set. Since there are only five employees, and the mode is different than the median, then the two highest earners must have the same salary. The salaries from smallest to largest is therefore:

?, ?, 34600, 35600, 35600

When the median gets the 3500 raise, the set becomes:

?, ?, 35600, 35600, 38100

So the new median is 35600.

The most common number is still 35600. So the mode hasn't changed: 35600.

4 0

3 years ago