Answer:
juhl;lnjk j bmnkj mkl
Step-by-step explanation:
Just factor to find the roots
<span>6x² - 6x - 72 = 0 </span>
<span>6(x² - x - 12) = 0 </span>
<span>6(x - 4)(x + 3) = 0 </span>
<span>B: x - 4</span>
Answer:
![(2^3)^2 = 64](https://tex.z-dn.net/?f=%282%5E3%29%5E2%20%3D%2064)
Step-by-step explanation:
Option 1:
Using the following rule:
![(a^n)^m = a^{nm}](https://tex.z-dn.net/?f=%28a%5En%29%5Em%20%3D%20a%5E%7Bnm%7D)
Put in our expression,
a = 2
n = 3
m = 2
![(a^n)^m = a^{nm}](https://tex.z-dn.net/?f=%28a%5En%29%5Em%20%3D%20a%5E%7Bnm%7D)
![(2^3)^2 = 2^{3*2}=2^6=64](https://tex.z-dn.net/?f=%282%5E3%29%5E2%20%3D%202%5E%7B3%2A2%7D%3D2%5E6%3D64)
Option 2:
Using the following rule:
![a^n * a^m = a^{n+m}](https://tex.z-dn.net/?f=a%5En%20%2A%20a%5Em%20%3D%20a%5E%7Bn%2Bm%7D)
Since our expression is the same as multiplying 2³ with itself, we can write it as a multiplication.
![(2^3)^2 = 2^3 * 2^3](https://tex.z-dn.net/?f=%282%5E3%29%5E2%20%3D%202%5E3%20%2A%202%5E3)
If we compare this with
, we can see that
a = 2
n = 3
m = 3 (in this case, n and m are equal)
![a^n*a^m = a^{n+m}](https://tex.z-dn.net/?f=a%5En%2Aa%5Em%20%3D%20a%5E%7Bn%2Bm%7D)
![2^3*2^3 = 2^{3+3} = 2^6 = 64](https://tex.z-dn.net/?f=2%5E3%2A2%5E3%20%3D%202%5E%7B3%2B3%7D%20%3D%202%5E6%20%3D%2064)
Answer: ![(2^3)^2 = 64](https://tex.z-dn.net/?f=%282%5E3%29%5E2%20%3D%2064)
The answer is E
step by step explanation: