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Math Help - Check my work please

  1. #1
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    Check my work please

    1. Using the laws of indices, remove the brackets from (x^3y^7)^4

    (a^mb^n)^k = a^{mk}b^{nk}

    (x^3y^7)^4 = x^{34}y^{74}

    2. Transpose the formula v = \sqrt{x+2y} for y

    v^2 = x+2y

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

    \frac{v^2}{2}-x=y

    3. Convert 18 degrees to radians.

    18 pi/180
    = pi/10r
    = 0.314 radians

    4. Simplify \frac{x^-3}{x^-2}
    x^{-3}{--2}
    x^{-3}{+2}
    x^-1

    5. Simplify cosAtanA
    tanA = \frac{SinA}{CosA}
    cosAtanA = cosA\frac{sinA}{cosA} = sinA

    6. Given that f(x) = x^2+1, write an expression for f(x+2)

    f(x) = x^2+1
    f(x+2) = (x+2)^2+1
    f(x+2) = x^2+4+1
    f(x+2)=x^2+5

    Thanks.
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  2. #2
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    Quote Originally Posted by gary223 View Post
    1. Using the laws of indices, remove the brackets from (x^3y^7)^4

    (a^mb^n)^k = a^{mk}b^{nk}

    (x^3y^7)^4 = x^{34}y^{74}

    2. Transpose the formula v = \sqrt{x+2y} for y

    v^2 = x+2y

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

    \frac{v^2}{2}-x=y


    6. Given that f(x) = x^2+1, write an expression for f(x+2)

    f(x) = x^2+1
    f(x+2) = (x+2)^2+1
    f(x+2) = x^2+4+1
    f(x+2)=x^2+5

    Thanks.
    Hi gary223,

    you have work to do for 1), 2) and 6)

    Q1 The notation means "multiply the indices"

    (2^2)^3=[(2)2][2(2)][2(2)]=2^{2(3)}=2^6

    (a^mb^n)^k=a^{m(k)}b^{n(k)}

    (x^3y^7)^4=x^{3(4)}y^{7(4)}=x^{12}y^{28}


    Q2 v^2=x+2y,\ v^2-x=2y,\ y=\frac{v^2-x}{2}

    Q6 (x+2)^2=(x+2)(x+2)=x(x+2)+2(x+2)=x^2+2x+2x+2(2)
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  3. #3
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    Hello gary223
    Quote Originally Posted by Archie Meade View Post
    Q6 (x+2)^2=(x+2)(x+2)=x(x+2)+2(x+2)=x^2+2x+2x+2(2)
    ... so f(x) = x^2 +4x + 4 + 1
    =x^2+4x+5
    Grandad
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  4. #4
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    Quote Originally Posted by Grandad View Post
    Hello gary223... so f(x) = x^2 +4x + 4 + 1
    =x^2+4x+5
    Grandad
    Quote Originally Posted by Archie Meade View Post
    Hi gary223,

    you have work to do for 1), 2) and 6)

    Q1 The notation means "multiply the indices"

    (2^2)^3=[(2)2][2(2)][2(2)]=2^{2(3)}=2^6

    (a^mb^n)^k=a^{m(k)}b^{n(k)}

    (x^3y^7)^4=x^{3(4)}y^{7(4)}=x^{12}y^{28}


    Q2 v^2=x+2y,\ v^2-x=2y,\ y=\frac{v^2-x}{2}

    Q6 (x+2)^2=(x+2)(x+2)=x(x+2)+2(x+2)=x^2+2x+2x+2(2)
    Thanks Grandad, I got the same answer.

    So x^{12}y^{28}

    and

    y=\frac{v^2-x}{2}

    are the correct answers or is there more to do?
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  5. #5
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    Hello gary223
    Quote Originally Posted by gary223 View Post
    Thanks Grandad, I got the same answer.

    So x^{12}y^{28}

    and

    y=\frac{v^2-x}{2}

    are the correct answers or is there more to do?
    No, they're complete.

    Grandad
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