8 kilometers is the answer
Hope this helps!:)
Change the x and y
Then solve for y
X= 8y+1
Subtract 1 from both sides
X-1=8y
Divide by 8
X-1 /8 = y
F(2) = 3(2)^2 + 2(2) + 4
= 3(4) + 4 + 4
= 12 + 8
f(2) = 20
f(a+h) = 3(a+h)^2 + 2(a+h) + 4
= 3(a^2 + 2ah + h^2) + 2a + 2h + 4
f(a+h) = 3a^2 + 6ah + 3h^2 + 2a + 2h + 4
Remark
If the lines are parallel then triangle RQS will be similar to triangle RTP
From that, all three lines in one triangle will bear the same ratio to all three lines of the second triangle.
Givens
PQ = 8
QR = 5
RS = 15
ST = x + 3
Ratio
QR/RP = RS/RT
Sub and solve
RP = 5 + 8
RP = 13
RT = 15 + x + 3
RT = 18 + x
5/13 = 15 / (18 + x) Cross multiply
5(18 + x) = 195 Remove the brackets on the left.
90 + 5x = 195 Subtract 90 from both sides.
5x = 105 Divide by 5
x = 105/5
x = 21 Answer <<<<<<<
