Given: h(t) = 25 - a·t²
h(0.5) = 21
Find: t such that h(t) = 0
Solution: h(0.5) = 25 - a·0.5² = 21
25 - 21 = a/4
4·4 = a = 16
Then
h(t) = 25 - 16t²
We want h(t) = 0, so
0 = 25 - 16t²
16t² = 25
t² = 25/16 = (5/4)²
t = 5/4 = 1.25
It takes 1.25 seconds for the entire 25 ft drop.
Answer:
Step-by-step explanation:
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Answer:
![x=2.4](https://tex.z-dn.net/?f=x%3D2.4)
Step-by-step explanation:
<u>Solving Equations Using Successive Approximations</u>
We need to find the solution to the equation
![f(x)=g(x)](https://tex.z-dn.net/?f=f%28x%29%3Dg%28x%29)
where
![f(x)=2x-5](https://tex.z-dn.net/?f=f%28x%29%3D2x-5)
![g(x)=x^2-6](https://tex.z-dn.net/?f=g%28x%29%3Dx%5E2-6)
The approximation has been already started and reached a state for x=2.5 where
![f(2.5)=0](https://tex.z-dn.net/?f=f%282.5%29%3D0)
![g(2.5)=2.5^2-6=0.25](https://tex.z-dn.net/?f=g%282.5%29%3D2.5%5E2-6%3D0.25)
The difference between the results is 0.25, we need further steps to reach a good solution (to the nearest tenth)
Let's test for x=2.4
![f(2.4)=-0.2](https://tex.z-dn.net/?f=f%282.4%29%3D-0.2)
![g(2.4)=2.4^2-6=-0.24](https://tex.z-dn.net/?f=g%282.4%29%3D2.4%5E2-6%3D-0.24)
The new difference is -0.2+0.24=0.04
It's accurate enough, thus the solution is
![\boxed{x=2.4}](https://tex.z-dn.net/?f=%5Cboxed%7Bx%3D2.4%7D)
Hello from MrBillDoesMath!
Answer:
(x+2) ( x^2 + 3)
Discussion:
x^3 + 2x^2 + 3x + 6 =
(
x^3 + 2x^2) + (3x + 6) = => factor "x^2" from the first term; factor 3
from the second term
x^2( x + 2) + 3( x+2) = => factor (x+2) from each term
(x+2) ( x^2 + 3)
Thank you,
MrB