B^n+5
When multiplying exponents the numbers are added
Answer:
4(2e - 3)(3e + 1)
Step-by-step explanation:
Given
24e² - 28e - 12 ← factor out 4 from each term
= 4(6e² - 7e - 3) ← factor the quadratic
Consider the factors of the product of the e² term and the constant term which sum to give the coefficient of the e- term.
product = 6 × - 3 = - 18 and sum = - 7
The factors are - 9 and + 2
Use these factors to split the e- term
6e² - 9e + 2e - 3 ( factor the first/second and third/fourth terms )
= 3e(2e - 3) + 1 (2e - 3) ← factor out (2e - 3) from each term
= (2e - 3)(3e + 1)
Then
24e² - 28e - 12 = 4(2e - 3)(3e + 1) ← in factored form
1. Use Difference of squares: a^2-b^2=(a+b)(a-b)
(x+k)(x-k)=0
2. Solve for x and k
x=+-k
k=+-x
Answer:
-60
Step-by-step explanation:
Distribute
6 x -3 = -18
6 x -7 = -42
-18 - 42 = -60
If my answer is right, please give it the brainliest.