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Math Help - Need help on Review

  1. #1
    Junior Member
    Joined
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    Need help on Review

    Can someone help with these problems I got a test on Thursday and I don't know how to do these. I can't use a cal by the way.

    1. What is tan(2π/3)?
    I know that tan is y/x so it is -1/2, sqroot 3/2 so how would I put this in an equation?

    2. If t^(2) = ln(1 + x) − ln(1 − x), what is x in terms of t?

    3. What are all solutions of the equation: sin^(2) x + 2 sin x = 5/4?

    4. If f (x) = √4x^(2) + 8, x
    ≥ 0, what is f ^(− 1) (x)?

    5. sin(2 cos^(−1) x) is equivalent to

    6. If f (x) = 1+sin x^(2) −sin x , −pie/2 ≤ x ≤ pie/2 , what is f^(−1) (x)?

    7. Whats equivalent to 4a/(2/b) + (1/2)

    8. √12 x Squareroot y/4

    9. What is the domain of f (x) = 2 − ln(x^(2) − 9)?

    10. A water tank is initially 3/5 full. After adding 5 gallons of water, it is 2/3 full. What is the capacity of the tank in gallons? I got 75 for this am I right? 3/5= 18/30, 2/3=20/30 and 2= 5 gallons so 2/30= 15x5=75
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  2. #2
    MHF Contributor
    Grandad's Avatar
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    South Coast of England
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    Hello goldenroll
    Quote Originally Posted by goldenroll View Post
    Can someone help with these problems I got a test on Thursday and I don't know how to do these. I can't use a cal by the way.

    1. What is tan(2π/3)?
    I know that tan is y/x so it is -1/2, sqroot 3/2 so how would I put this in an equation?
    Use \tan \pi/3 = \sqrt3, and then \tan 2\pi /3 = -\tan(\pi - \pi /3)

    2. If t^(2) = ln(1 + x) − ln(1 − x), what is x in terms of t?
    \ln(1+x) -\ln(1-x) = \ln\left(\frac{1+x}{1-x}\right) = t^2

    Raise e to the power of both sides:

    \frac{1+x}{1-x} = e^{t^2}

    Solve for x
    .

    3. What are all solutions of the equation: sin^(2) x + 2 sin x = 5/4?
    Write the equation as

    4\sin^2x+8\sin x - 5 =0

    This is a quadratic in \sin x. Factorise and solve.

    4. If f (x) = √4x^(2) + 8, x ≥ 0, what is f ^(− 1) (x)?
    I'm not quite sure what the square root sign covers here. I assume you mean:

    f(x) = \sqrt{4x^2 + 8}

    So, to find f^{-1}, let f(x)=y. Then

    y = \sqrt{4x^2 + 8}

    Now make x the subject of this equation, to give

    x = \sqrt{\frac{y^2-8}{4}}=f^{-1}(y)

    So f^{-1}(x) =
    \sqrt{\frac{x^2-8}{4}}

    5. sin(2 cos^(−1) x) is equivalent to
    Let \cos^{-1}x = y.

    \Rightarrow \cos y = x and \sin y = \sqrt{1-x^2}

    \Rightarrow \sin(2\cos^{-1}x) = \sin 2y = 2\sin y \cos y = ?


    Grandad
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