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 February 14th, 2010, 05:46 PM #1 Newbie   Joined: Feb 2010 Posts: 3 Thanks: 0 can someone help me solve this...simple 1. Let h(x)=3-2x and k(x)=-x^2+5x+8 Graph: R(x)=-2 I have no idea how to graph this. 2. What did I do wrong here? Find g(x+3)-g(3x) if g(x)=5x+4 5x+4+3-3(5x+4) 5x+7-15x-12 -10x-5 thanks in advance
February 14th, 2010, 08:14 PM   #2
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Re: can someone help me solve this...simple

Hello, Fi87!

Quote:
 1. Let $h(x)\:=\:3\,-\,2x\,\text{ and }\,k(x)\:=\:-x^2\,+\,5x\,+\,8$ Graph:[color=beige] .[/color]$R(x)\:=\:-2$ i have no idea how to graph this. [color=beige]. . [/color][color=blue]Neither do I . . . What in the world is R(x) ?[/color]

Quote:
 $2.\, \text{Given: }\:g(x) \:=\:5x\,+\,4$ $\text{Find: }\: g(x+3)\,-\,g(3x)$

Take it step by step . . .

$g(x + 3) \;=\;5(x + 3) \,+\,4 \;=\;5x\,+\,15 \,+\,4 \;=\;5x\,+\,19$

$g(3x) \;=\;5(3x)\,+\,4 \;=\;15x\,+\,4$

$\text{Subtract: }\;g(x+3)\,-\,g(3x) \;\;=\;\;(5x\,+\,19)\,-\,(15x\,+\,4) \;\;=\;\; 5x\,+\,19\,-\,15x\,-\,4 \;\;=\;\;-10x\,+\,15$

 February 14th, 2010, 08:40 PM #3 Newbie   Joined: Feb 2010 Posts: 3 Thanks: 0 Re: can someone help me solve this...simple Thanks for your help I really appreciate it! As for R(x) I don't know Hopefully someone who knows will come along.

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