The first step to solving this expression is to factor out the perfect cube
![\sqrt[3]{m^{2} n^{3} X n^{2} }](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7Bm%5E%7B2%7D%20%20n%5E%7B3%7D%20X%20n%5E%7B2%7D%20%20%20%7D%20)
The root of a product is equal to the product of the roots of each factor. This will make the expression look like the following:
![\sqrt[3]{ m^{2} n^{2} }](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B%20m%5E%7B2%7D%20n%5E%7B2%7D%20%20%7D%20)
Finally,, reduce the index of the radical and exponent with 3
n
![\sqrt[3]{ m^{2} n^{2} }](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B%20m%5E%7B2%7D%20n%5E%7B2%7D%20%20%7D%20)
This means that the correct answer to your question is n
![\sqrt[3]{ m^{2} n^{2} }](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B%20m%5E%7B2%7D%20n%5E%7B2%7D%20%7D%20)
.
Let me know if you have any further questions
:)
Answer:
0.926
Step-by-step explanation:
5/12 as a decimal is 0.416 and
5/9 as a decimal is 0.5 repeating
You just have to turn the fraction into a decimal and
add those together.
So your final equation would be 0.416+0.5
Answer:
-I haven't done this, but I have copied and pasted from another question and answer from the user syed514:
Graphing is one way to do the problem.But sometimes, graphing it is hard to do.So here’s an algebraic method.
If M(m1, m2) is the midpoint of two points A(x1, y1) and B(x2, y2),then m1 = (x1 + x2)/2 and m2 = (y1 + y2)/2.In other words, the x-coordinate of the midpointis the average of the x-coordinates of the two points,and the y-coordinate of the midpointis the average of the y-coordinates of the two points.
Let B have coordinates (x2, y2) in our problem.Then we have that 6 = (2 + x2)/2 and 8 = (3 + y2)/2.
Solving for the coordinates gives x2 = 10, y2 = 13
1. Take an arbitrary point that lies on the first line y=3x+10. Let x=0, then y=10 and point has coordinates (0,10).
2. Use formula
to find the distance from point
to the line Ax+By+C=0.
The second line has equation y=3x-20, that is 3x-y-20=0. By the previous formula the distance from the point (0,10) to the line 3x-y-20=0 is:
.
3. Since lines y=3x+10 and y=3x-20 are parallel, then the distance between these lines are the same as the distance from an arbitrary point from the first line to the second line.
Answer:
.