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Math Help - making the symbol the subject

  1. #1
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    making the symbol the subject

    Ok, again folks I am asking for your help!! I have managed to do the rest of the 100 questions but I am totally stuck on these ones (after I managed to get some great help on my other topic, thank you!)

    I need whatever is in brackets, to be the subject.

    (sq) = squared or whatever the value is

    Q1. v(sq) = 2gh (h)

    Q2. v = 5a
    -----
    a - b (b)

    Q3. c = [square root over this whole equation]2hr - h(sq) (r)

    Now these ones are simplifying again...

    Q1. a(sq) x a(3rd) x a(-5)

    Q2. [square root] 9a(4)b(sq)

    Q3. [square root] X(3rd) x X(5)

    Thank you very much for all your help!!
    Last edited by jimmybob; November 15th 2006 at 02:30 PM. Reason: missed a sum
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  2. #2
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by jimmybob View Post
    Q2. v = 5a
    -----
    a - b (b)
    I'll do this one, you need to do the other ones.

    You know: v=\frac{5a}{a-b}

    Multiply both sides by a-b to get: v(a-b)=5a

    Divide both sides by v to get: a-b=\frac{5a}{v}

    Subtract a from both sides to get: -b=\frac{5a}{v}-a

    Multiply both sides by -1 to get: \boxed{b=a-\frac{5a}{v}}

    If you want you can change that to: b=\frac{av}{v}-\frac{5a}{v}=\frac{av-5a}{v}=\frac{a(v-5)}{v}
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  3. #3
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    Quote Originally Posted by jimmybob View Post
    Ok, again folks I am asking for your help!! ...
    I need whatever is in brackets, to be the subject.
    (sq) = squared or whatever the value is

    Q1. v(sq) = 2gh (h)

    ...

    Q3. c = [square root over this whole equation]2hr - h(sq) (r)

    ...
    Hello Jimmybob,

    to Q1.:
    v^2=2gh\ \Longleftrightarrow \ h=\frac{v^2}{2g}

    to Q3.:
    c=\sqrt{2hr-h^2}. Square both sides:
    c^2=2hr-h^2. Add h^2 on both sides.
    c^2+h^2=2hr. Finally divide both sides by 2h:
    r=\frac{c^2+h^2}{2h}

    EB
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  4. #4
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    Quote Originally Posted by jimmybob View Post
    Ok, again folks I am asking for your help!...
    Q1. a(sq) x a(3rd) x a(-5)

    Q2. [square root] 9a(4)b(sq)

    Q3. [square root] X(3rd) x X(5)

    Thank you very much for all your help!!
    Hello, Jimmybob,

    to Q1.: The bases of all powers are the same therefore you only have to add the exponents:
    a^2 \cdot a^{\frac{1}{3}} \cdot a^{-5}=a^{2+\frac{1}{3}-5}=a^{-\frac{8}{3}}.

    to Q 2.: I've some trouble to read your problem. Therefore I'll offer you 2 different ways to this problem. You have to decide which result is the right one!
    \sqrt{9a^4} \cdot b^2 = 3a^2\cdot b^2
    Second edition:
    \sqrt{9a^4 \cdot b^2} = 3a^2\cdot b

    to Q 3.: Same procedure as the last question:
    \sqrt{x^{\frac{1}{3}}}\cdot x^5=x^{\frac{1}{6}}\cdot x^5 = x^{\frac{31}{6}}
    Second edition:
    \sqrt{x^{\frac{1}{3}}\cdot x^5}=\left(x^{\frac{1}{3}}\cdot x^5\right)^{\frac{1}{2}} = \left(x^{\frac{16}{3}}\right)^{\frac{1}{2}}=x^{\fr  ac{8}{3}}

    EB
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  5. #5
    Senior Member OReilly's Avatar
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    Quote Originally Posted by jimmybob View Post
    Now these ones are simplifying again...

    Q1. a(sq) x a(3rd) x a(-5)

    Q2. [square root] 9a(4)b(sq)

    Q3. [square root] X(3rd) x X(5)

    Thank you very much for all your help!!
    If I correctly interpreted your expressions which are not totally clear to see then is:

    Q1) a^2  \cdot a^3  \cdot a^{ - 5}  = a^5  \cdot \frac{1}{{a^5 }} = 1

    Q2) \sqrt {9a^4 b^2 }  = \sqrt 9 \sqrt {(a^2 )^2 } \sqrt {b^2 }  = 3a^2 b

    Q3) \sqrt {x^3  \cdot x^5 }  = \sqrt {x^8 }  = \sqrt {(x^4 )^2 }  = x^4
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