Need Help on Math..Algebra 2.?

Question

could you please help me on these questions. I can not figure them out...............I would very much appreciate it.
(*=multiply or times)

ok...so they are:

1. 2. simplify: (5^4/5 * 5^4/5) ^-10

4. simply: the square root of 64x^12

5. simplify: (7^3/4)^-2/9

6. write the expression in simplest form: the 3rd root (aka cubed root) of 1/5.

8. Let f(x)=16-x^2 and g(x)=4-x. Find (f * g)(x).

9. Let f(x)=4-x^2 and g(x)=2-x. Find f(g(x))


Those are the ones I just cant seem to understand..PLEASE help.............................

thank you.

Answer 1

1. 2. simplify: (5^4/5 * 5^4/5) ^-10
Inside the parentheses, since you have the same base, add the exponents. When you get done with that, you multiplythe exponent you found by the exponent outside the parentheses
(x^a*x^a)^b = (x^2a)^b = x^2ab

4. simply: the square root of 64x^12
Write the square root of the coefficient, then the square root of the variable.

5. simplify: (7^3/4)^-2/9
This is the same as the 2nd step in # 1.

6. write the expression in simplest form: the 3rd root (aka cubed root) of 1/5.
The simplest form is (1/5)^(1/3) or (1/5)^-(1/3).
The rationalized form is (1/5)(25)^(1/3)

8. Let f(x)=16-x^2 and g(x)=4-x. Find (f * g)(x).
Multiply the two functions.

9. Let f(x) = 4-x^2 and g(x)= 2-x. Find f(g(x))
Substitute g(x) into f(x)
f(g(x)) = 4 - (2-x)^2
Simplify.

Answer 2

In BODMAS, Multiplication is till now Addition besides, no desire for brackets. so the final 2 is accelerated with the help of 0 to get 0. Then there are fourteen '2's, making 28, and minus the two is 26. including the 0 for sure maintains to be at 26. Tada! XD