Answer:
5^20
Step-by-step explanation:
<u>L</u><u>a</u><u>w</u><u> </u><u>o</u><u>f</u><u> </u><u>E</u><u>x</u><u>p</u><u>o</u><u>n</u><u>e</u><u>n</u><u>t</u><u> </u><u>I</u>
Therefore:
<u>L</u><u>a</u><u>w</u><u> </u><u>o</u><u>f</u><u> </u><u>E</u><u>x</u><u>p</u><u>o</u><u>n</u><u>e</u><u>n</u><u>t</u><u> </u><u>I</u><u>I</u>
Thus:
First, you have to get the equation equal to 0. To do that, you need to subtract the
and 48 from the right side of the equation and put it on the left side. The new equation will read
TW is a radius of this circle.
|TW| = |WQ|
SV = VQ therefore m∠WVQ = m∠WVS = 90°. The triangle QVW is the rectangular triangle.
We calculate the length of WQ using the Pythagorean theorem.
Answer: TW = √58 ≈ 7.62
Move all terms that don't contain Y to the right side and solve.
The answer would be y=3-4x
For the parabola:
h(t) = -at2 + bt, the maximum occurs at -b/2a
For h(t) = -16t2 + 576t that is -576/(-2)(-16), or -576/32, or 18.
The projectile is at its highest point at 18 seconds.