The answer is 64 which goes into evenly 8 times 8, 49 7 times 7, 4, 2 times 2, 144 12 times 12
Answer:
False
Any integers that the numbers 5, 10 , 15 but 20 can be used as a counter argument against the statement.
Step-by-step explanation:
The claim is that A⊆B which stands for that A is a SUBSET in B, or that B contains A.
The truth is that B⊆A since 5 has more possible outcomes than 20 in the number of integers.
So the list of all possible answers are r5, r10, and r15 where N⊆Z.
For example I choose r=3 and r15, 3(15)= 45. I can use the number 45 as a counter argument that the statement of A⊆B is false.
Step-by-step explanation:
2x + 2y = 3 ( x2 for everything in the equation)
x - 4y = -1
4x+ 4y = 6
x - 4y = -1
Add the 1st line to the 2nd line to get rid of the y:
5x = 5
x - 4y = -1 ( you can pick any line from this system to be the second equation in this step, you can put 2x + 2y = 3 or 4x+ 4y = 6, anything above this step, but to make it simple chose the one you think is the easiest)
Solve for x, then plug the x value into the 2nd equation
x = 1
1 - 4y = -1
x = 1
-4y = -2 =) y = 1/2
x = 1
y= 1/2
Yes Sam is correct hope that helped would you like an explanation