Megan Homework: Difference between revisions

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Created page with '===True/False=== #<math>\sqrt{a} + \sqrt{a} = \sqrt{a+a}</math> #<math>\sqrt{8x} = 2 \sqrt{4x}</math>'
 
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#<math>\sqrt{a} + \sqrt{a} = \sqrt{a+a}</math>
#<math>\sqrt{a} + \sqrt{a} = \sqrt{a+a}</math>
#<math>\sqrt{8x} = 2 \sqrt{4x}</math>
#<math>\sqrt{8x} = 2 \sqrt{4x}</math>
#<math>e^{-1/2} = -\sqrt{e}</math>
# <math>(g \cdot f)(x) = g(x)f(x) </math>
# <math>(g \circ f)(x) = f(g(x)) </math>
#if <math>f(x) = 4x</math> and <math>g(x) = \sqrt{x} </math>
## <math>(g \circ f)(x) = 4 \sqrt{x} </math>
## <math>(g \circ g)(y) = \sqrt[4]{y} </math>
## <math>(g \circ f)(z) = (f \circ g)(z) </math>
#<math>t^2t^3 = t^6</math>
#<math>\frac{2x+1}{x-1} = \frac{2x}{x} = 2</math>

Revision as of 21:59, 12 April 2010

True/False

  1. [math]\displaystyle{ \sqrt{a} + \sqrt{a} = \sqrt{a+a} }[/math]
  2. [math]\displaystyle{ \sqrt{8x} = 2 \sqrt{4x} }[/math]
  3. [math]\displaystyle{ e^{-1/2} = -\sqrt{e} }[/math]
  4. [math]\displaystyle{ (g \cdot f)(x) = g(x)f(x) }[/math]
  5. [math]\displaystyle{ (g \circ f)(x) = f(g(x)) }[/math]
  6. if [math]\displaystyle{ f(x) = 4x }[/math] and [math]\displaystyle{ g(x) = \sqrt{x} }[/math]
    1. [math]\displaystyle{ (g \circ f)(x) = 4 \sqrt{x} }[/math]
    2. [math]\displaystyle{ (g \circ g)(y) = \sqrt[4]{y} }[/math]
    3. [math]\displaystyle{ (g \circ f)(z) = (f \circ g)(z) }[/math]
  7. [math]\displaystyle{ t^2t^3 = t^6 }[/math]
  8. [math]\displaystyle{ \frac{2x+1}{x-1} = \frac{2x}{x} = 2 }[/math]