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anonymous

  • one year ago

Solve the equation the square root of the quantity x minus 6 plus 2 equals 6 for the variable. Show each step of your solution process.

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  1. anonymous
    • one year ago
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    |dw:1437259693464:dw|

  2. Nnesha
    • one year ago
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    first combine like terms -6_2 and then take square both sides to cancel out square root

  3. Nnesha
    • one year ago
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    you can write square root as 1/2 root \[\huge\rm \sqrt{x} = x^\frac{ 1 }{ 2 }\] when you take square \[\huge\rm \sqrt{x} = (x^\frac{ 1 }{ 2 })^2\] both 2 cancel each othr out and then you x is left that's how you can take square to cancel ot square root :-)

  4. Owlcoffee
    • one year ago
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    \[\sqrt{x-6+2}=6\] The first step is alwas... ALWAYS trying to remove the square root since it is quite bothersome to deal with in algebra level. So, we will square both sides: \[(\sqrt{x-6+2})^2=6^2\] That'll cancel the square root and give us the following: \[x-6+2=6^2\] \[x-6+2=36\] And that is much simpler to solve, is it not?

  5. anonymous
    • one year ago
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    thank you both!! can you help me with one more??

  6. anonymous
    • one year ago
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    Let f(x) = 4x2 + x + 1 and g(x) = x2 – 2. Find g(f(x)). Show each step of your work.

  7. Owlcoffee
    • one year ago
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    That is what we call a "composite function", what you do is replace the function on all the x's of the other function, so if you have: \[f:f(x)=4x^2+x-1\] \[g:g(x)=x^2-2\] Then, in order for the transformation to occurr: \[g(x),f(x) \rightarrow (g,f): g(f(x)) \] We will have an equivalent of this: \[(g,f):g(f(x))=(f(x))^2-2\] So, therefore, g(f(x)) will be composed of: \[g(f(x))=(4x^2+x+1)^2-2\]

  8. anonymous
    • one year ago
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    then what..?

  9. Owlcoffee
    • one year ago
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    That's it.

  10. anonymous
    • one year ago
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    oh ok cool lol

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