Difference between revisions of "Megan Homework"

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