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Thread: 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; Nov 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: $\displaystyle v=\frac{5a}{a-b}$

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

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

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

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

    If you want you can change that to: $\displaystyle b=\frac{av}{v}-\frac{5a}{v}=\frac{av-5a}{v}=\frac{a(v-5)}{v}$
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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.:
    $\displaystyle v^2=2gh\ \Longleftrightarrow \ h=\frac{v^2}{2g}$

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

    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:
    $\displaystyle a^2 \cdot a^{\frac{1}{3}} \cdot a^{-5}=a^{2+\frac{1}{3}-5}=a^{-\frac{8}{3}}$.

    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!
    $\displaystyle \sqrt{9a^4} \cdot b^2 = 3a^2\cdot b^2$
    Second edition:
    $\displaystyle \sqrt{9a^4 \cdot b^2} = 3a^2\cdot b$

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

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

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

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