Answer:
The expressions which equivalent to
are:
⇒ B
⇒ C
Step-by-step explanation:
Let us revise some rules of exponent
Now let us find the equivalent expressions of 
A.
∵ 4 = 2 × 2
∴ 4 = 
∴
=
- By using the second rule above multiply 2 and (n + 2)
∵ 2(n + 2) = 2n + 4
∴
=
B.
∵ 4 = 2 × 2
∴ 4 = 2²
∴
= 2² ×
- By using the first rule rule add the exponents of 2
∵ 2 + n + 1 = n + 3
∴
=
C.
∵ 8 = 2 × 2 × 2
∴ 8 = 2³
∴
= 2³ ×
- By using the first rule rule add the exponents of 2
∵ 3 + n = n + 3
∴
=
D.
∵ 16 = 2 × 2 × 2 × 2
∴ 16 = 
∴
=
×
- By using the first rule rule add the exponents of 2
∵ 4 + n = n + 4
∴
=
E.
is in its simplest form
The expressions which equivalent to
are:
⇒ B
⇒ C
Answer: a. $2700
b. $10200
Step-by-step explanation:
a. The interest she would have earned at the end of the 6 years can be gotten using the formula
= PRT/100
= $7500 × 6% × 6
= $7500 × 6/100 × 6
= $7500 × 0.06 × 6
= $2700
b. Her balance when she wants to withdraw the money would be:
= $7500 + $2700
= $10200
Answer:

Step-by-step explanation:
The expression to transform is:
![(\sqrt[6]{x^5})^7](https://tex.z-dn.net/?f=%28%5Csqrt%5B6%5D%7Bx%5E5%7D%29%5E7)
Let's work first on the inside of the parenthesis.
Recall that the n-root of an expression can be written as a fractional exponent of the expression as follows:
![\sqrt[n]{a} = a^{\frac{1}{n}}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Ba%7D%20%3D%20a%5E%7B%5Cfrac%7B1%7D%7Bn%7D%7D)
Therefore ![\sqrt[6]{a} = a^{\frac{1}{6}}](https://tex.z-dn.net/?f=%5Csqrt%5B6%5D%7Ba%7D%20%3D%20a%5E%7B%5Cfrac%7B1%7D%7B6%7D%7D)
Now let's replace
with
which is the algebraic form we are given inside the 6th root:
![\sqrt[6]{x^5} = (x^5)^{\frac{1}{6}}](https://tex.z-dn.net/?f=%5Csqrt%5B6%5D%7Bx%5E5%7D%20%3D%20%28x%5E5%29%5E%7B%5Cfrac%7B1%7D%7B6%7D%7D)
Now use the property that tells us how to proceed when we have "exponent of an exponent":

Therefore we get: 
Finally remember that this expression was raised to the power 7, therefore:
[/tex]
An use again the property for the exponent of a exponent:
Answer:
4
Step-by-step explanation:
5 and 3 are two primary numbers for this question