# Algebra paper

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## Homework

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A.

xxx

xx find xxx xx intercept xx substitute x by 0

xxx y- xxxxxxxxx is

xxx

3.

(0,30) xx xxx xxxxx xx xxx third xxxxxx window

(45,0) is xxx point xxxxx xxx lase xxxx hits the ground

xxx

To xxxx the height 30 feet away xxxx xxx face of xxx xxxxxxxxx

xxxxxxxxxxx

xxxxxxxxxxx

xx xxxxxx xxx xxxxx on xxx xxxxx xx xxxx thirty xxxxx xx xxx right xxxx xxx origin xxx xxxx xxxx up xxxxx xx intersect xxx graph at xxx xxxxx desired xxx then xxxx the corresponding y-value of this xxxxxxx

xxx

xxx xxxxxxxx xxxxxxxx is xxx xxxxx xxxxxxxxxx

The graph xx a xxxxxxxxxx representation xx the path of xxx xxxxx xxxx xxxxxx the first xxxxxxxx xxxx xxxxxxx xxxxxx xxxx the xxxxx xxx xx xxxxxxxx meaning.

# _1391705847.unknown

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A.

xx

𝑦 = − xx

3 𝑥 x xx

xx xxxx xxx x- intercept xx xxxxxxxxxx x xx x

0 = − 2

xx𝑥 x 30

2

3 𝑥 x 30

𝑥 xxxxx

2 = 45

The xx xxxxxxxxx is

𝑦 = − xx

xx 0 x xx x 30

xx

3.

xxxxxx xx xxx point xx xxx xxxxx xxxxxx xxxxxx

xxxxxx is xxx point where xxx xxxx beam xxxx xxx ground

xx

To find xxx height 30 feet xxxx from the face xx xxx xxxxxxxx

Y=-2/3x+30

xxxxxxxxxx

xx xxxxxx the xxxxx on the xxxxx xx xxxx xxxxxx xxxxx to the right from the origin xxx xxxx xxxx

xx until we xxxxxxxxx the graph xx the point desired xxx then read the corresponding y-value xx

xxxx point.

5.

The relevant quadrant xx the xxxxx quadrant.

xxx graph is x xxxxxxxxxx representation xx xxx xxxx of xxx xxxxx beam xxxxxx xxx xxxxx quadrant

only xxxxxxx beyond xxxx the xxxxx has xx xxxxxxxx xxxxxxxx

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