To the person who makes a graph and doesn't copy from another question
1 answer:
Answer:
H=9;8)
B=(5;4) (ball)
R=(7;0) (hit point)
B'=symetric of B axis perpendicular of x in R
B'=(7+(7-5);4)=(9;4)
Equation BR: y-4=(0-4)/(7-5)(x-5)==>y=-2x+14
Equation RB': y-4=(4-0)/(9-7)(x-9)==>y=2x-14
Is H a point of RB'? y=2x-14 : 8=? 2*9-14==>8=?4 No!
you will not make your putt
Step-by-step explanation:
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