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Math Help - [SOLVED] last 3 homework problems im stumpped on

  1. #1
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    [SOLVED] last 3 homework problems im stumpped on

    1. They want me to evaluate the expression using the values given in table for (fog)(4)

    x | 1,5,11,12
    f(x)|-4, 11, 3, 15

    x |-5, -4, 1, 4
    g(x)|1, -7, 5, 11


    2.Solve the equation
    4^7-3x=1/16 (I know this has something to do with logs, but i dont know what)

    3.While traveling in a car, the centrifugal force a passenger experiences as the car drives in a circle varies jointly as the mass of the passenger and the square of the speed of the car. If the passenger experiences a force of 352.8 newtons when the car is moving at a speed of 70 kph and the passenger has a mass of 80 kg, find the force a passenger experiences when the car is moving at 20kph and the passenger has a mass of 90kg.

    Im not asking for the answers, i just dont know how to solve the problems. The formulas, steps, or functions are what I am looking for.
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  2. #2
    Senior Member Pinkk's Avatar
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    Although I'm not too sure on what the last two problems are asking for, I can help with the first problem.

    When you see something like (f\circ g)(4), what essentially is being asked is the evaluation of f(g(4)). So find the evaluation of g(4) and then find the evaluation of f(x) at the value x=g(4).
    Last edited by Pinkk; March 21st 2009 at 06:22 PM.
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  3. #3
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    Quote Originally Posted by Pinkk View Post
    Although I'm not too sure on what the last two problems are asking for, I can help with the first problem.

    When you see something like (f*g)(4), what essentially is being asked is the evaluation of f(g(4)). So find the evaluation of g(4) and then find the evaluation of f(x) at the value x=g(4).
    i dont think im following. How do you evaluate it from a table? What im thinking is that in the table of g(x), when g is 11, x=4
    Is that what you're refering to?
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  4. #4
    Senior Member Pinkk's Avatar
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    When looking at the chart, first look at what g(x) is when x=4. In the chart, g(4)=11. Now you look at what f(x) is when x=11, and in the chart, f(11)=3. Notice how g(4)=11, so what you have is (f\circ g)(4)=f(g(4))=f(11)=3.
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  5. #5
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    Hello, ninobrn99!

    Using the values in the tables, evaluate: . (f\circ g)(4)

    . . \begin{array}{c|cccc}<br />
x & 1 &5 &11 & 12 \\ \hline<br />
f(x)& \text{-}4 & 11 & 3 & 15\end{array} \qquad\qquad \begin{array}{c|cccc}x & \text{-}5&\text{-}4 & 1 & 4 \\ \hline<br />
g(x)& 1 & \text{-}7 & 5 & 11 \end{array}
    (f\circ g)(4) means: . f(g(4))

    From the second table: . g(4) = 11

    . . So: . f(g(4)) \;=\;f(11)

    From the first table: . f(11) = 3

    . . Therefore: . f(g(4)) \:=\:3



    2. Solve the equation: . 4^{7-3x}\:=\:\frac{1}{16}

    We have: . 4^{7-3x} \:=\:\frac{1}{4^2} \quad\Rightarrow\quad 4^{7-3x} \:=\:4^{-2}

    Hence: . 7-3x \:=\:-2 \quad\Rightarrow\quad -3x \:=\:-9 \quad\Rightarrow\quad x \:=\:3

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  6. #6
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    Okay, thank you both. After seeing it, I was on the right path, i just couldn't figure out where I was hanging up. Thank you...now just the troublesome word problem
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  7. #7
    Senior Member Pinkk's Avatar
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    No problem. I suggest getting used to the [ math ] tagging with problems, because I thought your second problem was 4^{7} - 3x = \frac{1}{16}, not 4^{7-3x}=\frac{1}{16}
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  8. #8
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    Quote Originally Posted by Pinkk View Post
    No problem. I suggest getting used to the [ math ] tagging with problems, because I thought your second problem was 4^{7} - 3x = \frac{1}{16}, not 4^{7-3x}=\frac{1}{16}
    sorry, i thought the problem was
    the same as you because the -3x didnt look like a superscript. My bust on that and yes, i am practicing on my Latex. Thanks again.
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  9. #9
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    okay, i think i figured out the word problem

    okay...so the formula is
    <br />
F=\frac{mv^2}{r}<br />
    with that, I need to figure out what r is in order to figure out the second part of the problem.
    <br />
352.8=\frac{(80)(70)^2}{r}<br />
    so now i try to solve for r:
    <br />
352.8r=(80)(4900)<br />

    <br />
352.8r=392000<br />
    <br />
r=1111.11<br />

    With r figured out, i can now solve the following:
    <br />
F=\frac{(90)(20)^2}{1111.11}<br />

    <br />
F=32.4 newtons<br />

    I know the latex is all messed up, but does that look right?
    Last edited by ninobrn99; March 23rd 2009 at 08:48 AM.
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