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Math Help - a lot of problems

  1. #1
    soldier
    Guest

    a lot of problems

    help would be greatly appreciated.

    z is jointly proportional to x and y, z=3 when y is 5 and x is 10. Find z when y is 2 and x is 4.

    y is inversely proportional to the square of x, y is 8 when x is 3. Find y when x is 5.

    The intensity I of light received from a source varies inversely as the square of the distance d from the source. If the light intensity is a 5 foot-candles at 17 feet, find the intensity at 16 feet.

    find the inverse of the function y = 2x(squared) -5

    Verify if the functions f = (square root) x-9 and g = x (squared) + 9 are inverses

    Given f(x)=2x-1 and g(x) = x(squared) + x + 2

    a. find (f-g)(x)

    b. find f(g(x))

    c. Find g (f(2))
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  2. #2
    Senior Member DivideBy0's Avatar
    Joined
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    From
    Melbourne, Australia
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    Quote Originally Posted by soldier View Post
    help would be greatly appreciated.

    For the first question,
    z is jointly proportional to x and y, z=3 when y is 5 and x is 10. Find z when y is 2 and x is 4.

    y is inversely proportional to the square of x, y is 8 when x is 3. Find y when x is 5.

    The intensity I of light received from a source varies inversely as the square of the distance d from the source. If the light intensity is a 5 foot-candles at 17 feet, find the intensity at 16 feet.

    find the inverse of the function y = 2x(squared) -5

    Verify if the functions f = (square root) x-9 and g = x (squared) + 9 are inverses

    Given f(x)=2x-1 and g(x) = x(squared) + x + 2

    a. find (f-g)(x)

    b. find f(g(x))

    c. Find g (f(2))
    z \propto xy \implies z=kxy for some k.

    If z = 3 when x = 10 and y = 5,

    3=k \cdot 10 \cdot 5

    k=\frac{3}{50}

    \implies z=\frac{3}{50}xy

    When y = 2 and x = 4,

    z=\frac{3}{50} \cdot 2 \cdot 4

    z=\frac{24}{50}=\frac{12}{25}

    For the second question,

    y \propto \frac{1}{x^2} \implies y=\frac{k}{x^2} for some k.

    If y = 8 when x = 3,

    8=\frac{k}{3^2}

    k=72

    \implies y=\frac{72}{x^2}

    When x = 5,

    y=\frac{72}{5^2}

    y=\frac{72}{25}

    For the third question,

    I \propto \frac{1}{d^2} \implies I=\frac{k}{d^2}

    If I = 5 when d = 17,

    5=\frac{k}{17^2}

    k=1445

    \implies I = \frac{1445}{d^2}

    If d = 16,

    I=\frac{1445}{16^2}

    I=\frac{1445}{256}

    For question 4, y=f(x)=2x^2-5. To find the inverse, swap around the y and x and solve for y.

    x=2y^2-5

    y^2=\frac{x+5}{2}

    y=\pm \sqrt{\frac{x+5}{2}}

    So f^{-1}(x)=\sqrt{\frac{x+5}{2}} or f^{-1}(x)=-\sqrt{\frac{x+5}{2}}

    For question 5, if f(x) and g(x) are inverse funtions, then g(x)=f^{-1}(x). Also, f^{-1}(f(x))=x, so if f(g(x))=x, then we know they are inverses.

    f(g(x))=\sqrt{(x^2+9)-9}=\sqrt{x^2}=x, so they are inverses.

    For question 6,

    a. (f-g)(x)=f(x)-g(x)=2x-1-(x^2+x+2)=-x^2+x-3

    b. f(g(x))=2(x^2+x+2)-1=2x^2+2x+4-1=2x^2+2+3

    c. g(f(2))=(2(2)-1)^2+(2(2)-1)-3=3^2+3-3=9
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