Answer:
The ray is EF in this diagram ...
Answer:
what is your question
Step-by-step explanation:
Answer:
309 feet
Step-by-step explanation:
given that the height can be represented by a quadratic equation, we can say that the general form of the equation will look something like this:
h(x) = Ax² + Bx + C
we know that at the starting point of the launch that time = 0 (i.e x = 0) and hence h (x) = 0, if we substitute this into our equation, we can find the value for C
h(0) = A(0)² + B(0) + C = 0
C = 0
Hence the equation becomes
h(x) = Ax² + Bx
Given when x = 1, h(1) = 121,
121 = A(1)² + B(1)
A + B = 121 ------> eq 1
Given when x = 2, h(2) = 224,
224 = A(2)² + B(2)
4A + 2B = 224 (divide both sides by 2)
2A + B = 112------> eq 2
Solving the system of equations which comprise eq 1 and eq 2 using your favorite method, we end up with A = -9 and B = 130
our equation becomes:
h(x) = -9x² + 130x
when x = 3
h(3) = -9(3)² + 130(3) = 309 feet
Is there supposed to be a question or a picture with this ?
Answer:
Step-by-step explanation:
If u = 6 then substitute so
6(6+6)=6 know solve but this isn't going to be correct because 6 plus 6 is 12 times 6 isn't 3
So your answer isn't correct. I think it would be a negative number or decimal because 6 is larger than 3. Hope this helps!
it is u(u+6)=3
I got u=6