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Thread: 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
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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))
    $\displaystyle z \propto xy \implies z=kxy$ for some k.

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

    $\displaystyle 3=k \cdot 10 \cdot 5$

    $\displaystyle k=\frac{3}{50}$

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

    When y = 2 and x = 4,

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

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

    For the second question,

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

    If y = 8 when x = 3,

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

    $\displaystyle k=72$

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

    When x = 5,

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

    $\displaystyle y=\frac{72}{25}$

    For the third question,

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

    If I = 5 when d = 17,

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

    $\displaystyle k=1445$

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

    If d = 16,

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

    $\displaystyle I=\frac{1445}{256}$

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

    $\displaystyle x=2y^2-5$

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

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

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

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

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

    For question 6,

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

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

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