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:
Answer:
p < - 15
Step-by-step explanation:
Given
< - 3 ( multiply both sides by 5 to clear the fraction )
p < - 15
X-intercept: y = 0
y-intercept: x = 0
3x + 4y = 0
x-intercept: subtitute y = 0
3x + 4 · 0 = 0
3x = 0
x = 0 → (0; 0)
y-intercept: subtitute x = 0
3 · 0 + 4y = 0
4y = 0
y = 0 → (0; 0)
Answer:
its not possible
Step-by-step explanation:
