Answer:
![\large\boxed{2.\ ab^{-3x}=a\left(\dfrac{1}{b}\right)^{3x}=a\left[\left(\dfrac{1}{b}\right)^3\right]^x}](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7B2.%5C%20ab%5E%7B-3x%7D%3Da%5Cleft%28%5Cdfrac%7B1%7D%7Bb%7D%5Cright%29%5E%7B3x%7D%3Da%5Cleft%5B%5Cleft%28%5Cdfrac%7B1%7D%7Bb%7D%5Cright%29%5E3%5Cright%5D%5Ex%7D)
Step-by-step explanation:
![Use:\ a^{-n}=\left(\dfrac{1}{a}\right)^n\\------------\\\\(4)^{-3x^2}=\left[(4)^{-1}\right]^{3x^2}=\left(\dfrac{1}{4}\right)^{3x^2}](https://tex.z-dn.net/?f=Use%3A%5C%20a%5E%7B-n%7D%3D%5Cleft%28%5Cdfrac%7B1%7D%7Ba%7D%5Cright%29%5En%5C%5C------------%5C%5C%5C%5C%284%29%5E%7B-3x%5E2%7D%3D%5Cleft%5B%284%29%5E%7B-1%7D%5Cright%5D%5E%7B3x%5E2%7D%3D%5Cleft%28%5Cdfrac%7B1%7D%7B4%7D%5Cright%29%5E%7B3x%5E2%7D)
![Use:\ a^{-n}=\left(\dfrac{1}{a}\right)^n\ and\ (a^n)^m=a^{nm}\\--------------------\\\\ab^{-3x}=a\cdot b^{-3x}=a\left[(b)^{-1}\right]^{3x}=a\left(\dfrac{1}{b}\right)^{3x}\\\\ab^{-3x}=a\left(\dfrac{1}{b}\right)^{3x}=a\left[\left(\dfrac{1}{b}\right)^3\right]^x](https://tex.z-dn.net/?f=Use%3A%5C%20a%5E%7B-n%7D%3D%5Cleft%28%5Cdfrac%7B1%7D%7Ba%7D%5Cright%29%5En%5C%20and%5C%20%28a%5En%29%5Em%3Da%5E%7Bnm%7D%5C%5C--------------------%5C%5C%5C%5Cab%5E%7B-3x%7D%3Da%5Ccdot%20b%5E%7B-3x%7D%3Da%5Cleft%5B%28b%29%5E%7B-1%7D%5Cright%5D%5E%7B3x%7D%3Da%5Cleft%28%5Cdfrac%7B1%7D%7Bb%7D%5Cright%29%5E%7B3x%7D%5C%5C%5C%5Cab%5E%7B-3x%7D%3Da%5Cleft%28%5Cdfrac%7B1%7D%7Bb%7D%5Cright%29%5E%7B3x%7D%3Da%5Cleft%5B%5Cleft%28%5Cdfrac%7B1%7D%7Bb%7D%5Cright%29%5E3%5Cright%5D%5Ex)
Answer:
I'll answer one question
Step-by-step explanation:
a. x >= -4. This reads, the domain is greater than and equal to negative four to infinity.
The solution of the equation x2+7x is 0, -7 using the quadratic formula.
Step-by-step explanation:
x2 + 7x = 0
-b ± √
b2 - 4(ac)/ 2a
substitution,
a = 1, b = 7, c = 0
= -7 ± √(7)2 - 4(1 x 0) / 2 x 1
= - 7
± 7 / 2
x = 0 , -7
I would try Khan Academy. He does very good explainations
