Which type of line of symmetry does the figure have? <br>

Which expression is equivalent to ( 4mn / m^-2n^6)^-2

inn [45]

Here are a few rules with exponents that apply here:

Raising a power to a power: $(x^m)^n=x^{m*n}$
Dividing exponents of the same base: $\frac{x^m}{x^n}=x^{m-n}$
Converting negative exponents to positive ones: $x^{-m}=\frac{1}{x^m}\ \textsf{and}\ \frac{1}{x^{-m}}=x^m$

Firstly, solve the outer exponent:

$(\frac{4mn}{m^{-2}n^{6}})^{-2}=\frac{4^{-2}m^{-2}n^{-2}}{m^{-2*-2}n^{6*-2}}=\frac{4^{-2}m^{-2}n^{-2}}{m^4n^{-12}}$

Next, divide:

$\frac{4^{-2}m^{-2}n^{-2}}{m^4n^{-12}}=4^{-2}m^{-2-4}n^{-2-(-12)}=4^{-2}m^{-6}n^{10}$

Next, convert the negative exponents:

$4^{-2}m^{-6}n^{10}=\frac{n^{10}}{4^2m^6}=\frac{n^{10}}{16m^6}$

<u>Your final answer is n^10/16m^6 , or B.</u>

8 0

3 years ago

Hi guys, Can anyone help me with this tripple integral? Thank you:)

OleMash [197]

I don't usually do calculus on Brainly and I'm pretty rusty but this looked interesting.

We have to turn K into the limits of integration on our integrals.

Clearly 0 is the lower limit for all three of x, y and z.

Now we have to incorporate

x+y+z ≤ 1

Let's do the outer integral over x. It can go the full range from 0 to 1 without violating the constraint. So the upper limit on the outer integral is 1.

Next integral is over y. y ≤ 1-x-z. We haven't worried about z yet; we have to conservatively consider it zero here for the full range of y. So the upper limit on the middle integration is 1-x, the maximum possible value of y given x.

Similarly the inner integral goes from z=0 to z=1-x-y

We've transformed our integral into the more tractable

$\displaystyle \int_0^1 \int_0^{1-x} \int _0^{1-x-y} (x^2-z^2)dz \; dy \; dx$

For the inner integral we get to treat x like a constant.

$\displaystyle \int _0^{1-x-y} (x^2-z^2)dz = (x^2z - z^3/3)\bigg|_{z=0}^{z= 1-x-y}=x^2(1-x-y) - (1-x-y)^3/3$

Let's expand that as a polynomial in y for the next integration,

$= y^3/3 +(x-1) y^2 + (2x+1)y -(2x^3+1)/3$

The middle integration is

$\displaystyle \int_0^{1-x} ( y^3/3 +(x-1) y^2 + (2x+1)y -(2x^3+1)/3)dy$

$= y^4/12 + (x-1)y^3/3+ (2x+1)y^2/2- (2x^3+1)y/3 \bigg|_{y=0}^{y=1-x}$

$= (1-x)^4/12 + (x-1)(1-x)^3/3+ (2x+1)(1-x)^2/2- (2x^3+1)(1-x)/3$

Expanding, that's

$=\frac{1}{12}(5 x^4 + 16 x^3 - 36 x^2 + 16 x - 1)$

so our outer integral is

$\displaystyle \int_0^1 \frac{1}{12}(5 x^4 + 16 x^3 - 36 x^2 + 16 x - 1) dx$

That one's easy enough that we can skip some steps; we'll integrate and plug in x=1 at the same time for our answer (the x=0 part doesn't contribute).

$= (5/5 + 16/4 - 36/3 + 16/2 - 1)/12$

$=0$

That's a surprise. You might want to check it.

Answer: 0

6 0

3 years ago

A rectangle has a total area of 1116 square feet. If the length of the rectangle is 12 ft, what is the width of the rectangle?

Pani-rosa [81]

the width of the rectangle is 93

5 0

3 years ago

Read 2 more answers

What is the rule when finding the sum of numbers from 1-100?

Sphinxa [80]

Answer:

Step-by-step explanation:

Sum of 1 through 100 = 100×(1+100)/2

6 0

3 years ago

Help with problems 1 & 3

vlabodo [156]

One side of the rectangle is x=2, the other side is 2x-5
Add up all the four sides: (x+2) +(x+2)+(2x-5)+(2x-5)=54
6x-6=54
x=10

#3: suppose the first integer is x, then the second one is x+2
x(x+2)=255
x^2 +2x -255 =0
factor the quadratic equation: (x+17)(x-15)=0
x=-17, which is impossible, or x=15
so the two positive integers are 15 and 17

5 0

3 years ago

Which type of line of symmetry does the figure have? ​

Which type of line of symmetry does the figure have?