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Thread: Multiple Choice Question

  1. #1
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    Multiple Choice Question

    Hi
    I few question i am having trouble with:
    1)P and Q lie on the curve $\displaystyle y=x^3$. The x coordinates of P and Q are 2 and (2+h) respectively. The gradient of chord PQ is:
    A.$\displaystyle \frac{h^3-8}{h-2}$
    B.$\displaystyle 12+6h$
    C.12
    D.$\displaystyle \frac{(2+h)^3-8}{h}$
    E.$\displaystyle 12+6h+h^2$

    2)The equation $\displaystyle logx=y(log3)+1$ is equivalent to the equation:
    A.$\displaystyle x=10.3^y$
    B.$\displaystyle x=30^y$
    C.$\displaystyle x=3^y+10$
    D.$\displaystyle x=y^3+10$
    E.$\displaystyle x=10y^3$

    3)If the angle between the lines 2y=8x+10 and 3x-6y=22 is $\displaystyle \theta$, then $\displaystyle tan\theta$ is best approximated by:
    A)1.17 B)1.40 C)2 D)0.86 E)1

    P.S
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  2. #2
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    Quote Originally Posted by Paymemoney View Post
    Hi
    I few question i am having trouble with:
    1)P and Q lie on the curve $\displaystyle y=x^3$. The x coordinates of P and Q are 2 and (2+h) respectively. The gradient of chord PQ is:
    A.$\displaystyle \frac{h^3-8}{h-2}$
    B.$\displaystyle 12+6h$
    C.12
    D.$\displaystyle \frac{(2+h)^3-8}{h}$
    E.$\displaystyle 12+6h+h^2$

    2)The equation $\displaystyle logx=y(log3)+1$ is equivalent to the equation:
    A.$\displaystyle x=10.3^y$
    B.$\displaystyle x=30^y$
    C.$\displaystyle x=3^y+10$
    D.$\displaystyle x=y^3+10$
    E.$\displaystyle x=10y^3$

    3)If the angle between the lines 2y=8x+10 and 3x-6y=22 is $\displaystyle \theta$, then $\displaystyle tan\theta$ is best approximated by:
    A)1.17 B)1.40 C)2 D)0.86 E)1

    P.S
    You should only ask one question in a thread in the future

    1. P = (2,8) and Q= ([2+h], [2+h]^3)

    The gradient, m, is equal to $\displaystyle \frac{\Delta y}{\Delta x} = \frac{(2+h)^3-8}{(2+h-2)} $

    You don't even need to expand the cubic to see the answer

    ======================================

    2. $\displaystyle x = 10^{ylog_{10}3+1}$. Use the laws of exponents to simplify
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  3. #3
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    ok this is what i got for question 2

    $\displaystyle x=10log3^y+1$

    $\displaystyle 10log = 1$

    $\displaystyle 3^y+10^1$

    $\displaystyle x=3^y+10$

    Answer says its $\displaystyle x=10.3^y$
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  4. #4
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    Quote Originally Posted by Paymemoney View Post
    ok this is what i got for question 2

    $\displaystyle x=10log3^y+1$

    $\displaystyle 10log = 1$ <--- where did you get this from?

    $\displaystyle 3^y+10^1$

    $\displaystyle x=3^y+10$

    Answer says its $\displaystyle x=10.3^y$
    The way I would have done it is to say that $\displaystyle c^ac^b = c^{a+b}$

    $\displaystyle 10^{log3^y+1} = 10^{log3^y} \cdot 10^1 = 10 \cdot 3^y$
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  5. #5
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    Quote Originally Posted by Paymemoney View Post
    ok this is what i got for question 2

    $\displaystyle x=10log3^y+1$
    You mean $\displaystyle x= 10^{log 3^y+ 1}$ and that is equal to $\displaystyle 10^1(10^{log 3^y})= 10(3^y)$

    $\displaystyle 10log = 1$

    $\displaystyle 3^y+10^1$

    $\displaystyle x=3^y+10$

    Answer says its $\displaystyle x=10.3^y$
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