Answer:
What exactly is your question?
Step-by-step explanation:
Answer:
110 cm²
Step-by-step explanation:
The area of a triangle is 1/2 times base times height.
A = 1/2bh
A = 1/2(22)(10)
A = 110
Hello! Lets solve this :)
So for the first question :
h(t) = -16t^2 + 150
h(0)= -16 (0)^2 + 150
h(0)= -16(0) + 150
h(0)= 0 + 150
h(0)= 150 feet
Second question
We need to find here the time when the ball hit the ground; or when the height h(t) is 0.
In equation form
h(t) --> 0=16t^2 + 150
16t^2 = 150
t^2=150/16
t= <span>√150/16
t= 5</span><span>√6/4
t= 5 (2.45)/4
t= 3.06 seconds
</span>
Hope this helps! If you have any other questions or would like further explanation just let me know! :)
Answer:
<u>
</u>
Step-by-step explanation:
The given question is :
8 = 4 * 2
32 = 16 * 2
∴
= 
=
Answer:
Step-by-step explanation:
We can write 
as follows:
![\frac{11s}{s^2-12s+52}\\=11\left [ \frac{s}{s^2-12s+52} \right ]\\=11\left [ \frac{s}{(s-6)^2+16} \right ]\\=11\left [ \frac{s-6+6}{(s-6)^2+16} \right ]\\=11\left [ \frac{s-6}{(s-6)^2+16} \right ]+\frac{66}{(s-6)^2+16}](https://tex.z-dn.net/?f=%5Cfrac%7B11s%7D%7Bs%5E2-12s%2B52%7D%5C%5C%3D11%5Cleft%20%5B%20%5Cfrac%7Bs%7D%7Bs%5E2-12s%2B52%7D%20%5Cright%20%5D%5C%5C%3D11%5Cleft%20%5B%20%5Cfrac%7Bs%7D%7B%28s-6%29%5E2%2B16%7D%20%5Cright%20%5D%5C%5C%3D11%5Cleft%20%5B%20%5Cfrac%7Bs-6%2B6%7D%7B%28s-6%29%5E2%2B16%7D%20%5Cright%20%5D%5C%5C%3D11%5Cleft%20%5B%20%5Cfrac%7Bs-6%7D%7B%28s-6%29%5E2%2B16%7D%20%5Cright%20%5D%2B%5Cfrac%7B66%7D%7B%28s-6%29%5E2%2B16%7D)
To find:
![L^{-1}\left [ \frac{11s}{s^2-12s+52 \right ]}\\=L^{-1}\left [ 11\left [ \frac{s-6}{(s-6)^2+16} \right ]+\frac{66}{(s-6)^2+16} \right ]](https://tex.z-dn.net/?f=L%5E%7B-1%7D%5Cleft%20%5B%20%5Cfrac%7B11s%7D%7Bs%5E2-12s%2B52%20%5Cright%20%5D%7D%5C%5C%3DL%5E%7B-1%7D%5Cleft%20%5B%2011%5Cleft%20%5B%20%5Cfrac%7Bs-6%7D%7B%28s-6%29%5E2%2B16%7D%20%5Cright%20%5D%2B%5Cfrac%7B66%7D%7B%28s-6%29%5E2%2B16%7D%20%5Cright%20%5D)
We will use formulae:
we get solution as :
![L^{-1}\left [ 11\left [ \frac{s-6}{(s-6)^2+16} \right ]+\frac{66}{(s-6)^2+16} \right ]\\=L^{-1}\left [ 11\left [ \frac{s-6}{(s-6)^2+4^2} \right ]+\frac{66}{4}\left [ \frac{4}{(s-6)^2+4^2} \right ] \right ]\\=11e^{6t}\cos 4t+\frac{33}{2}e^{6t}\sin 4t](https://tex.z-dn.net/?f=L%5E%7B-1%7D%5Cleft%20%5B%2011%5Cleft%20%5B%20%5Cfrac%7Bs-6%7D%7B%28s-6%29%5E2%2B16%7D%20%5Cright%20%5D%2B%5Cfrac%7B66%7D%7B%28s-6%29%5E2%2B16%7D%20%5Cright%20%5D%5C%5C%3DL%5E%7B-1%7D%5Cleft%20%5B%2011%5Cleft%20%5B%20%5Cfrac%7Bs-6%7D%7B%28s-6%29%5E2%2B4%5E2%7D%20%5Cright%20%5D%2B%5Cfrac%7B66%7D%7B4%7D%5Cleft%20%5B%20%5Cfrac%7B4%7D%7B%28s-6%29%5E2%2B4%5E2%7D%20%5Cright%20%5D%20%5Cright%20%5D%5C%5C%3D11e%5E%7B6t%7D%5Ccos%204t%2B%5Cfrac%7B33%7D%7B2%7De%5E%7B6t%7D%5Csin%204t)
