Answer:
m=2 and n=3
Step-by-step explanation:
<u>Step</u> :-
Given ![[ 2 x^{n}y^{2} ]^m = 4 x^6 y^4](https://tex.z-dn.net/?f=%5B%202%20x%5E%7Bn%7Dy%5E%7B2%7D%20%5D%5Em%20%3D%204%20x%5E6%20y%5E4)
using algebraic formula 
now
now equating 'x' powers, we get
....(1)
now
Equating 'y' powers ,we get
2 m=4
m=2
substitute m= 2 in equation (1)
we get
2 n=6
n=3
verification:-
substitute m=2 and n=3 , we get
both are equating so m= 2 and n=3
Answer:
x=-27, y=-15
Step-by-step explanation:
In the attached file
Answer:
A perfect square is a whole number that is the square of another whole number.
n*n = N
where n and N are whole numbers.
Now, "a perfect square ends with the same two digits".
This can be really trivial.
For example, if we take the number 10, and we square it, we will have:
10*10 = 100
The last two digits of 100 are zeros, so it ends with the same two digits.
Now, if now we take:
100*100 = 10,000
10,000 is also a perfect square, and the two last digits are zeros again.
So we can see a pattern here, we can go forever with this:
1,000^2 = 1,000,000
10,000^2 = 100,000,000
etc...
So we can find infinite perfect squares that end with the same two digits.
I believe it is 13/60
First, you would make them have common denominators
4/5 (12/12) is 48/60
7/12 (5/55 is 35/60
Second, you want to find the difference
48-35=13
So from that, I would assume that the difference between the two is 13/60
The answer to that is (the product)
