Answer:
x = ![12\sqrt{2}](https://tex.z-dn.net/?f=12%5Csqrt%7B2%7D)
Step-by-step explanation:
Cos 45= 12/x
x=12/Cos 45
![x=\frac{12}{\frac{\sqrt{2}}{2}}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B12%7D%7B%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%7D)
![x=\frac{24}{\sqrt{2}}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B24%7D%7B%5Csqrt%7B2%7D%7D)
x= ![12\sqrt{2}](https://tex.z-dn.net/?f=12%5Csqrt%7B2%7D)
NASA launches a rocket at t = 0 seconds. Its height, in meters above sea-level, as a function of time is given by
![h(t)=-4.9t^2+121t+283](https://tex.z-dn.net/?f=h%28t%29%3D-4.9t%5E2%2B121t%2B283)
The sea level is represented by h = 0, therefore, to find the corresponding time when h splashes into the ocean we have to solve for t the following equation:
![h(t)=0\implies-4.9t^2+121t+283=0](https://tex.z-dn.net/?f=h%28t%29%3D0%5Cimplies-4.9t%5E2%2B121t%2B283%3D0)
Using the quadratic formula, the solution for our problem is
![\begin{gathered} t=\frac{-(121)\pm\sqrt{(121)^2-4(-4.9)(283)}}{2(-4.9)} \\ =\frac{121\pm142.083778...}{9.8} \\ =\frac{121+142.083778...}{9.8} \\ =26.8452834694... \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20t%3D%5Cfrac%7B-%28121%29%5Cpm%5Csqrt%7B%28121%29%5E2-4%28-4.9%29%28283%29%7D%7D%7B2%28-4.9%29%7D%20%5C%5C%20%3D%5Cfrac%7B121%5Cpm142.083778...%7D%7B9.8%7D%20%5C%5C%20%3D%5Cfrac%7B121%2B142.083778...%7D%7B9.8%7D%20%5C%5C%20%3D26.8452834694...%20%5Cend%7Bgathered%7D)
The rocket splashes after 26.845 seconds.
The maximum of this function happens at the root of the derivative. Differentiating our function, we have
![h^{\prime}(t)=-9.8t+121](https://tex.z-dn.net/?f=h%5E%7B%5Cprime%7D%28t%29%3D-9.8t%2B121)
The root is
![-9.8t+121=0\implies t=\frac{121}{9.8}=12.347](https://tex.z-dn.net/?f=-9.8t%2B121%3D0%5Cimplies%20t%3D%5Cfrac%7B121%7D%7B9.8%7D%3D12.347)
Then, the maximum height is
![h(12.347)=1029.99](https://tex.z-dn.net/?f=h%2812.347%29%3D1029.99)
1029.99 meters above sea level.
4x - 2y = 10
-2y = -4x + 10
y = 4/2x - 10/2
y = 2x - 5 <==