(not sure )
my method
let projection of b onto line dq be M
BM is 2r
then let the remaining angle ( right) be K
in triangle BMK ,
MK = 29-5-r
By Pyth. theorem ,
23^2 = (2r)^2+(29-5-r)^2
529=4r^2+576-48r+r^2
5r^2-48r+47=0
r = 8.493237063 or 1.106762937(rejected)
Answer:
It should be "one solution"
Step-by-step explanation:
After graphing the equations, the two lines only intersect at one point which makes it "one solution." Hope this helps.
Ok
y=-0.04x^2+8.3x+4.3
when the rocket reaches the ground (when height=0, ie when y=0), then the rocket will land, find the x coordinate
set y=0
0=-0.04x^2+8.3x+4.3
use quadratic formula
if you have ax^2+bx+c=0, then
x=

a=-0.04
b=8.3
c=4.3
x=

x=208.017 or -0.516785
xrepresents horizontal distance
you cannot have a negative horizontal distance unless you fired and theh wind blew it backwards
therefor x=280.017 is the answer
208.02 m
I would describe it as talented and accurate. I have not gotten anything lower than a b+ in any of my classes, including math.