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:
Let the width be x.
Length is 8 feet more than width. Length = x + 8
Area = x(x + 8)
width increased by 4, that is, (x + 4)
Length decreased by 5, (x + 8 - 5) = (x + 3)
Area = (x + 4)(x +3)
Area remains the same
x(x + 8) = (x+4)(x +3)
x² + 8x = x(x +3) + 4(x +3)
x² + 8x = x² +3x + 4x +12
x² + 8x = x² +7x +12 Eliminate x² from both sides
8x = 7x + 12
8x - 7x = 12
x = 12
Dimensions of original rectangle : x, x + 8
12, 12 +8 = 12, 20
Original rectangle is 20 feet by 12 feet
The three points give you three equations, which you can solve by your favorite method.
.. 0m +0n +b = 6
.. 1m +0n +b = -3
.. 0m +2n +b = 5
(m, n, b) = (-9, -1/2, 6)
so the equation of the plane can be written as
.. z = -9x -1/2y +6
_____
In standard form, this would be
.. 18x +y +2z = 12
Let's solve your equation step-by-step.
-2/3p+ 1/6= 7/10
Answer:
p= -4/5