Answer:
She should add 10x to both sides
Step-by-step explanation:
Answer:
D: {-20, -13, 1, 6}
R: {-20, -8, 11, 13}
Step-by-step explanation:
Given the relation, {(–20, 11), (6, –8), (1, –20), (–13, 13)}, all x-values (inputs) make up the domain of the relation while all y-values make up the range of the relation.
Therefore:
Domain: {-20, -13, 1, 6}
Range: {-20, -8, 11, 13}
Answer:
I was at the store getting groceries I saw that there was a sell on bags of candy The original price was 4.00 dollars The sale is 1/2 off a bag of candy what is the new price. The answer is 2.00 dollars.
First we'll do two basic steps. Step 1 is to subtract 18 from both sides. After that, divide both sides by 2 to get x^2 all by itself. Let's do those two steps now
2x^2+18 = 10
2x^2+18-18 = 10-18 <<--- step 1
2x^2 = -8
(2x^2)/2 = -8/2 <<--- step 2
x^2 = -4
At this point, it should be fairly clear there are no solutions. How can we tell? By remembering that x^2 is never negative as long as x is real.
Using the rule that negative times negative is a positive value, it is impossible to square a real numbered value and get a negative result.
For example
2^2 = 2*2 = 4
8^2 = 8*8 = 64
(-10)^2 = (-10)*(-10) = 100
(-14)^2 = (-14)*(-14) = 196
No matter what value we pick, the result is positive. The only exception is that 0^2 = 0 is neither positive nor negative.
So x^2 = -4 has no real solutions. Taking the square root of both sides leads to
x^2 = -4
sqrt(x^2) = sqrt(-4)
|x| = sqrt(4)*sqrt(-1)
|x| = 2*i
x = 2i or x = -2i
which are complex non-real values
