Results 1 to 5 of 5

Thread: Check my work please

  1. #1
    Junior Member
    Joined
    Aug 2007
    Posts
    37

    Check my work please

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

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

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

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

    $\displaystyle v^2 = x+2y$

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

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

    3. Convert 18 degrees to radians.

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

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

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

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

    $\displaystyle f(x) = x^2+1$
    $\displaystyle f(x+2) = (x+2)^2+1$
    $\displaystyle f(x+2) = x^2+4+1$
    $\displaystyle f(x+2)=x^2+5$

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    4
    Quote Originally Posted by gary223 View Post
    1. Using the laws of indices, remove the brackets from $\displaystyle (x^3y^7)^4$

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

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

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

    $\displaystyle v^2 = x+2y$

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

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


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

    $\displaystyle f(x) = x^2+1$
    $\displaystyle f(x+2) = (x+2)^2+1$
    $\displaystyle f(x+2) = x^2+4+1$
    $\displaystyle f(x+2)=x^2+5$

    Thanks.
    Hi gary223,

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

    Q1 The notation means "multiply the indices"

    $\displaystyle (2^2)^3=[(2)2][2(2)][2(2)]=2^{2(3)}=2^6$

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

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


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

    Q6 $\displaystyle (x+2)^2=(x+2)(x+2)=x(x+2)+2(x+2)=x^2+2x+2x+2(2)$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello gary223
    Quote Originally Posted by Archie Meade View Post
    Q6 $\displaystyle (x+2)^2=(x+2)(x+2)=x(x+2)+2(x+2)=x^2+2x+2x+2(2)$
    ... so $\displaystyle f(x) = x^2 +4x + 4 + 1$
    $\displaystyle =x^2+4x+5$
    Grandad
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Aug 2007
    Posts
    37
    Quote Originally Posted by Grandad View Post
    Hello gary223... so $\displaystyle f(x) = x^2 +4x + 4 + 1$
    $\displaystyle =x^2+4x+5$
    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"

    $\displaystyle (2^2)^3=[(2)2][2(2)][2(2)]=2^{2(3)}=2^6$

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

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


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

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

    So $\displaystyle x^{12}y^{28}$

    and

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

    are the correct answers or is there more to do?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello gary223
    Quote Originally Posted by gary223 View Post
    Thanks Grandad, I got the same answer.

    So $\displaystyle x^{12}y^{28}$

    and

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

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

    Grandad
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: Mar 12th 2011, 04:05 AM
  2. Work--please check my work for me
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Feb 1st 2011, 09:02 AM
  3. Replies: 1
    Last Post: Jul 21st 2010, 06:02 PM
  4. check my work please
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Jan 20th 2007, 07:11 PM
  5. can someone check my work thanks!!
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: Jan 3rd 2007, 10:58 AM

Search Tags


/mathhelpforum @mathhelpforum