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MATH133_U2_answer

MATH133_U2_answer - 2 A by looking at the graph if my...

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NAME: _Jennifer Harper Student Answer form Unit 2; show all work for full credit; trial and error (guess and check) methods are not appropriate; use methods covered in class 1. A). x^2-6x-16=0 x^2-6x-16=0 (x+2)(x-8)=0 X+2=0 or x-8=0 X=-2 or x=8 B). 6x^2+3x-18=0 -b+-srt(b^2-4ac)/2a A=6 B=3 C=-18 -3+-srt([9-((4)*(6)*(-18))]/12 [-3+-srt(441)]/12 -3+-21/12 -3+21/12 and -3-21/12 X=1.5 and -2 C. discriminant is 3^2-4(6)(-18)=144
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Unformatted text preview: 2. A) by looking at the graph if my numbers are correct the points will cross at the x-axis. . B) yes it has a maximum. It is at the peak of the parabola. C) The coordinates of the vertex are 2,9. D) The line of symmetry is x=2. 3. A).p(x)=-0.4x2+60x-1500 B).p(50)=-0.4(50)2+60(50)-1500 =-1000+3000-1500 p(50)=500 C). A profit of about 95 would need to be sold. X=-60-34.6/-0.8=94.6 D).95...
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