Results 1 to 14 of 14

Thread: i am lousy at math~ plz help mi solve!

  1. #1
    Junior Member
    Joined
    Jan 2006
    Posts
    38

    i am lousy at math~ plz help mi solve!

    [IMG]D:\Documents and Settings\fang\My Documents\Scan0002.jpg[/IMG]


    OPS~ how to put picture?!?!?!?!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    5
    Quote Originally Posted by xiaoz
    [IMG]D:\Documents and Settings\fang\My Documents\Scan0002.jpg[/IMG]


    OPS~ how to put picture?!?!?!?!
    When creating a post scroll down past the text editor window,
    you will see a button marked manage attachments, click on
    it and it will allow you to upload a jpg attachment.

    RonL
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2006
    Posts
    38
    kk. thxz .. here is questions i dunno.. i am confuse at math tis yr!!
    Attached Thumbnails Attached Thumbnails i am lousy at math~ plz help mi solve!-scan0002.jpg  
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jan 2006
    Posts
    38
    they say canot use calculator.. i dunno how to do the working.. my math fail badly too.. HELP!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    5
    1.a Evaluate:

    $\displaystyle \sqrt{1\frac{11}{25}}$

    first change what is under the root symbol from a mixed number to an
    improper fraction:

    $\displaystyle
    \sqrt{1\frac{11}{25}}=\sqrt{\frac{36}{25}}=\frac{ \sqrt {36}}{\sqrt{25}}
    $

    You should now be able to complete this without a calculator as
    both $\displaystyle 36$ and $\displaystyle 25$ are perfect squares.

    RonL
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Jan 2006
    Posts
    38
    o.. i find out the answer liao... is 6 over 5.. thx!!

    1b.. explain plz..
    Last edited by xiaoz; Jun 6th 2006 at 09:51 PM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    5
    1.b Evaluate $\displaystyle \sqrt{2120}$ without a calculator given that

    $\displaystyle \sqrt{2.12}=1.456$ and $\displaystyle \sqrt{21.2}=4.004$.

    We want to write $\displaystyle 2120$ as a product of a square and one of
    the number we are given the square root of. Now:

    $\displaystyle 2120=1000 \times 2.12$,

    and

    $\displaystyle 2120=100 \times 21.2$.

    $\displaystyle 1000$ is not the square of an integer, but $\displaystyle 100=10^2$. So we can write:

    $\displaystyle
    \sqrt{2120}=\sqrt{100 \times 21.2}=\sqrt{100}\times \sqrt{21.2}
    $.

    RonL
    Last edited by CaptainBlack; Jun 6th 2006 at 11:22 PM.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    5
    1.c Evaluate without a calculator:

    $\displaystyle
    \left(-\frac{1}{216}\right)^{-2/3}
    $

    Now:

    $\displaystyle
    \left(-\frac{1}{216}\right)^{-2/3}=\frac{1}{(-\frac{1}{216})^{2/3}}
    $$\displaystyle
    =\frac{1}{((-\frac{1}{216})^{1/3})^2}=\frac{1}{((-\frac{1}{(216)^{1/3}}))^2}
    $

    Now this can be evaluated from the innermost brackets outward, as 216 is
    a perfect cube (of a small number which you can find by trial and error).

    RonL
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello,xiaoz!

    Here's #6 . . .

    6(a) Solve the inequality: $\displaystyle 5 \;< \;6x - 13 \;< \;7$
    We have: .$\displaystyle 5 \;< \; 6x - 13 \;< \;7$

    Add 13 to each side: .$\displaystyle 18 \:< \:6x \:< \:20$

    Divide by 6: . $\displaystyle 3 \;< \;x \;< \;\frac{10}{3}$

    (b) Find a fraction $\displaystyle x$ such that: $\displaystyle \frac{7}{11}\;<\:x\:<\:\frac{8}{11}$
    There are several ways to solve this problem.

    (1) To find a number between two given numbers, average the numbers.
    . . . $\displaystyle x\;=\;\frac{\frac{7}{11} + \frac{8}{11}}{2} \;=\;\frac{15}{22}$

    (2) Convert to decimals: $\displaystyle \begin{array}{cc}\frac{7}{11} = 0.63\hdots \\ \frac{8}{11} = 0.72\cdots\end{array}$
    Then let $\displaystyle x$ be any decimal between $\displaystyle 0.64$ and $\displaystyle 0.72$

    For example: . $\displaystyle x\:=\:0.65\:=\:\frac{65}{100} \:= \:\frac{13}{20}$

    (c) If $\displaystyle x^{2y} = 5$, find the value of $\displaystyle 2x^{6y} - 4$
    Change the first expression into the second, step by step.

    We have: .$\displaystyle x^{2y}\:=\:5$

    Cube both sides: .$\displaystyle (x^{2y})^3\:=\:5^3\quad\Rightarrow\quad x^{6y}\:=\:125$

    Multiply by 2: .$\displaystyle 2x^{6y}\;=\;250$

    Subtract 4: .$\displaystyle 2x^{6y} - 4 \:= \:246$
    Last edited by Soroban; Jun 7th 2006 at 03:52 AM.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member
    Joined
    Jan 2006
    Posts
    38

    Unhappy

    thxz.. 1b i undersyand.. 1c i dunno still... 6c the last part.. the mutiply by 2 de.. got the power still can mutiply?? how about question 2a and b.. explaint plz.. thxz
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, xiaoz!

    Let me give 1(c) a try . . .

    1(c) Simplify: $\displaystyle \left(-\frac{1}{216}\right)^{-\frac{2}{3}}$
    Did you know that a negative exponent on a fraction flips the fraction?
    . . $\displaystyle \left(\frac{a}{b}\right)^{-n}\;=\;\left(\frac{b}{a}\right)^n$

    We have: $\displaystyle \left(-\frac{1}{216}\right)^{-\frac{2}{3}}\:=\: (-216)^{\frac{2}{3}} $


    We know that: $\displaystyle -216 = (-6)^3$

    So we have: $\displaystyle [(-6)^3]^{\frac{2}{3}}\;=\;6^{(3\cdot\frac{2}{3})} \;=\;6^2\;=\;36$
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Junior Member
    Joined
    Jan 2006
    Posts
    38

    Red face

    woow..i did not know about tt formula.. hahax.. thxz to u.. i understand liao..
    other question.. plz help.. explain to mi..
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Member Ranger SVO's Avatar
    Joined
    Apr 2006
    From
    Abilene Tx
    Posts
    90
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Junior Member
    Joined
    Jan 2006
    Posts
    38

    Unhappy

    kk.. thxz all...now plz help mi wif question 2 , 3 , 4 , 5
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: Mar 26th 2011, 06:05 AM
  2. math homework that i just can't solve.
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: May 4th 2010, 04:35 PM
  3. 3 math questions. please solve
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Nov 26th 2008, 04:06 AM
  4. Math Help. Please solve
    Posted in the Trigonometry Forum
    Replies: 0
    Last Post: Nov 22nd 2008, 07:51 AM
  5. pls help solve this math problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Mar 25th 2008, 05:47 AM

Search Tags


/mathhelpforum @mathhelpforum