Need Help on Math. (algebra 2) PLEASE HELP?

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

First, this is a qns on indices. You have to know the following rules of indices:

a) a^x *a^b = a^(x+b)
b) (a^x) / (a^b) = a^ (x-b)
c) (a^x)^b = a^(bx)

Thus,
(5^4/5 * 5^4/5) ^-10
=5^(4/5+4/5)^-10 making use of rule (a)
=5^(8/5)^-10
=5^[8/5 * (-10)] making use of rule (c)
=5^(-16)

Qns 5. can be done similiarly, using the 3 rules shown above. Try it for yourself. The answer you should get is 7^(-1/6).

As for qns 4, simply: the square root of 64x^12.
You can treat the number 64 and the variable x^12 separately, so since 64=8*8 and x^12=x^6*x^6
√(64x^12) = √(8*8 *x^6*x^6)
= 8*x^6
= 8x^6

I'm not sure if you have typed qns 6 correctly. Perhaps it is (1/5)^3 instead? But if your qns is correct, then it make require the use of rationalisation.
3√(1/5) = 1/(3√5)
= 1/(3√5) * (√5/√5)
= (√5)/(15)
Rationalisation means to get rid of the surd or in other words the root in the denominatior of the fraction.

As for qns 8 and 9, both are essentially asking for the same thing as (f * g)(x)= f(g(x)).
I shall show you 8 and you can go try 9 yourself.
8. Let f(x)=16-x² and g(x)=4-x. Find (f * g)(x).

(f * g)(x)=fg(x)
Read from the the function closes to x, in this case is it g, which is just beside the x in fg(x). If it is gf(x), then you have to subsitute f(x) and then leave g outside.
=f(4-x)
substitute into the function f with (4-x) in place of x.
=16 - (4-x)² expand
= 16 - (16 - 8x + x²)
= 16 - 16 + 8x - x²
= 8x - x²

Answer 2

1.2.
(5^4/5 * 5^4/5)^-10
= (5^(4/5+4/5))^-10
= (5^8/5)^-10
= 5^(-16)
= 1/(5^16)

4.
sqrt(64x^12)
= (64x^12)^(1/2)
= 64^(1/2) * (x^12)^(1/2)
= 8x^6

5.
(7^3/4)^-2/9
= 7^(-6/36)
= 7^(-1/6)
= 1/7^(1/6)

6.
3rd root of 1/5
= (1/5)^(1/3)

8.
f(x) = 16-x^2
g(x) = 4-x

(f*g)(x)
= (16-x^2)(4-x)
= 64 - 4x^2 - 16x + x^3

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

f(g(x))
= 4 - (g(x))^2
= 4 - (2 - x)^2
= 4 - (4 + x^2 - 4x)
= -x^2 + 4x

Answer 3

**** observe: when you consider which you have distinctive that aces do count style as being under 9, I quite are transforming into rid of all my references to the possibility that Aces do no longer count style.***** threat is expressed as a fragment "favorable/attainable." for each subject, "attainable" is the style of procedures of drawing 2 taking part in cards out of fifty two (which one we draw first does not remember). We call that fifty two-choose-2, and it is equivalent to (fifty two! / (2! *(fifty two-2)!)) it quite is comparable to fifty two*fifty one / 2, or, 1326 So there are 1326 attainable information on the thank you to entice 2 taking part in cards. What approximately favorable? nicely, when you consider that there are 13 hearts, the style of information on the thank you to entice 2 hearts is 13-choose-2. it is equivalent to (13! / (2! *(13-2)!)) or, seventy 8. subsequently, the threat of drawing 2 hearts is seventy 8 / 1326, or 39/663, that's a pair of 6% hazard. for 2 numbers below 9, we would desire to nicely known what number there are in the deck. If Aces count style (as "one million"), then there are 32 such taking part in cards. So the favorable outcomes are 32-choose-2. (32! / (2! *(32-2)!)) = 496 So the threat is 496/1326 = 248/663. it quite is a pair of 37% hazard. playstation fifty two! is fifty two factorial and it skill fifty two * fifty one * 50......* 2 * one million **** Ah, so do you prefer to nicely known the possibility of having the two of those outcomes? Many have advised you to characteristic the two possibilities mutually. yet to accomplish that could be to overlook with regard to the shown fact that many of the suited outcomes are the two 2 hearts AND 2 numbers below 9! what number outcomes overlap? nicely, there are 8 hearts under 9. So the information on the thank you to entice 2 of them are 8-choose-2. it quite is (8! / (2! *(8-2)!)), or 28. to detect the threat of the two effect, we count style favorable outcomes like this: style of information on the thank you to entice 2 hearts PLUS information on the thank you to entice 2 numbers below 9 MINUS style of procedures that overlap. So: seventy 8 + 496 - 28, or 546 favorable outcomes. The threat is 546/1326, or ninety one/221, that's a pair of forty-one% hazard, no longer forty 3%.

Answer 4

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