SOLUTION: I received help yesterday(11/11) on these problems and think I understand. I have submitted my attempts at some to see if I have the right idea and a few that still escape me. Some

Algebra ->  Algebra  -> Polynomials-and-rational-expressions -> SOLUTION: I received help yesterday(11/11) on these problems and think I understand. I have submitted my attempts at some to see if I have the right idea and a few that still escape me. Some      Log On

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Question 61000: I received help yesterday(11/11) on these problems and think I understand. I have submitted my attempts at some to see if I have the right idea and a few that still escape me. Some seem pretty straight forward now but others elude me. Thanks again for such timely help.
For the following, letf(x)=x+5; let g(x)=x^2^-2; let h(x)=-3x
for h{f(4)} I set 4=x, the value of f(4) then being 9; the expression h(9) would then be -27???right???
for g(x+3), I set (x)=x+3 and then the value of g(x+3) being x^3^+3x^2^-2x-6???
I lost it on the following and need help:
1. Same parameters (f-g)(x)
2. h(2x^2^+4x)
The last problem has nothing to do with the parameters above. It has just given me fits.
3. 5x^3/2^-25=0
Thanks in advance for your help and this great program.
Answer by funmath(2873) About Me  (Show Source):
You can put this solution on YOUR website!
For the following, letf(x)=x+5; let g(x)=x^2-2; let h(x)=-3x
for h(f(4)) I set 4=x, the value of f(4) then being 9; the expression h(9) would then be -27???right???RIGHT!!!!
:
for g(x+3), I set (x)=x+3 and then the value of g(x+3)
g(x+3)=(x+3)^2-2
g(x+3)=(x+3)(x+3)-2
g(x+3)=x^2+3x+3x+9-2
g(x+3)=x^2+6x+7
:
1. Same parameters (f-g)(x)
(f-g)(x)=(x+5)-(x^2-2)
(f-g)(x)=x+5-x^2+2
(f-g)(x)=-x^2+x+7
:
2. h(2x^2+4x)=-3(2x^2+4x)
h(2x^2+4x)=-6x^2-12x
:
I had trouble interpretting this one. From what I can tell you are putting ^ before AND after your exponents. (You only need it before, and if you have a fractional exponent, put it in parentheses.) If this isn't what you meant, let me know.
3. 5x^(3/2)-25=0
5x^(3/2)-25+25=0+25
5x^(3/2)=25
5x%5E%283%2F2%29%2F5=25%2F5
x%5E%283%2F2%29=5
%28x%5E%283%2F2%29%29%5E%282%2F3%29=5%5E%282%2F3%29
x%5E1=5%5E%282%2F3%29
highlight%28x=5%5E%282%2F3%29%29
:
Check by substitution to see if it's true.
5%285%5E%282%2F3%29%29%5E%283%2F2%29-25=0
5%285%5E%286%2F6%29%29-25=0
5%285%29-25=0
25-25=0
0=0 we're right!!!
Happy Calculating!!!