The path of a firecracker is modeled by the equation h(t) = -t^2 + 19t + 14, where h(t) is the hight, in feet, of the firecracke
r at any given time, t. What is the hight of the firecracker after 4 seconds from launch.
A. 63 feet
B. 68 feet
C. 70 feet
D. 74 feet
A. 63 feet
B. 68 feet
C. 70 feet
D. 74 feet
1 answer:
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Answer:
given you are asked to simplify
Step-by-step explanation:
You have to multiply the numerator and denominator by the denominator's conjugate.
The conjugate of a+bi is a-bi.
When you multiply conjugates, you just have to multiply first and last.
(a+bi)(a-bi)
a^2-abi+abi-b^2i^2
a^2+0 -b^2(-1)
a^2+-b^2(-1)
a^2+b^2
See no need to use the whole foil method; the middle terms cancel.
So we are multiplying top and bottom of your fraction by (-3+4i):
So you will have to use the complete foil method for the numerator. Let's do that:
(-3+5i)(-3+4i)
First: (-3)(-3)=9
Outer:: (-3)(4i)=-12i
Inner: (5i)(-3)=-15i
Last: (5i)(4i)=20i^2=20(-1)=-20
--------------------------------------------Combine like terms:
9-20-12i-15i
Simplify:
-11-27i
Now the bottom (-3-4i)(-3+4i):
F(OI)L (we are skipping OI)
First:-3(-3)=9
Last: -4i(4i)=-16i^2=-16(-1)=16
---------------------------------------------Combine like terms:
9+16=25
So our answer is