First write the equation
3/n = 5
Multiply both sides by n
3/n * n = 5 * n
Simplify
3 = 5 * n
Divide both sides by 5
3/5 = 5 * n/5
Simplify
3/5 = n
Divide
0.6 = n
So 3/0.6 = 5
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
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I believe it's 15, but correct me if I'm wrong
This is because there is one 10 in 15. This gives you 3 tens and 5 in the ones.
