Let x,y be two different numbers
suppose x^2=y^2
then x^2-y^2=0
which yields (x+y)(x-y)=0
so either x=y or x=-y
In any case, x and y must be the same value
also when a vairable is squared like y=x^2
we must note that there are 2 possible solutions
x=(+/-)sqrt(y)
Answer:
If two powers have the same base then we can multiply the powers. When we multiply two powers we add their exponents.
Step-by-step explanation:
Answer:
Perpendicular
Step-by-step explanation:
m=y2-y1/x2-x1
m=15-3/14-(-2)
m=3/4
A. Alright, we want to multiply one equation by a constant to make it cancel out with the second. Since the first equation has a "blank" y, let's multiply the first equation by <em>2</em>.
3x-y=0 → 2(3x-y=0) = 6x - 2y = 0
5x+2y=22
The answer for this part would be: 6x - 2y = 0 and 5x + 2y = 22
B. So now we combine them:
6x - 2y = 0
+ + +
5x + 2y = 22
= = =
11x + 0 = 22 ← The answer
C. Now that we have the equation 11x = 22, we solve for x
11x = 22 ← Divide both sides by 11
x = 2 ← The answer
D. Now that we have x=2, we plug that back in to 5x+2y=22 and solve for y:
5(2)+2y = 22
10 + 2y = 22
2y = 12
y = 6
<u>Therefore, the solution to this problem is x = 2 and y = 6</u>
Well think about it
50 pages = 1 hour (60 mins)
so it would take laine 4 hours in total to read 200 pages at the rate
hope that helps :)
