Take $1.75 from the $10 then dived the 55cents into the sum. Thus $10 - $1.75 = $8.25. / .55 =15 miles exactly
Yes. I will help you with anything
Answer:
2.83 s
Step-by-step explanation:
We are given that
![h(t)=-16t^2+10t+100](https://tex.z-dn.net/?f=h%28t%29%3D-16t%5E2%2B10t%2B100)
When a rock reached the ground then h(t)=0
![-16t^2+10t+100=0](https://tex.z-dn.net/?f=-16t%5E2%2B10t%2B100%3D0)
![16t^2-10t-100=0](https://tex.z-dn.net/?f=16t%5E2-10t-100%3D0)
Using quadratic formula
![x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-b%5Cpm%5Csqrt%7Bb%5E2-4ac%7D%7D%7B2a%7D)
![t=\frac{10\pm\sqrt{(-10)^2-4\times 16\times (-100)}}{2(16)}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B10%5Cpm%5Csqrt%7B%28-10%29%5E2-4%5Ctimes%2016%5Ctimes%20%28-100%29%7D%7D%7B2%2816%29%7D)
![t=\frac{10\pm\sqrt{100+6400}}{32}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B10%5Cpm%5Csqrt%7B100%2B6400%7D%7D%7B32%7D)
![t=\frac{10+\sqrt{6500}}{32}=2.83](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B10%2B%5Csqrt%7B6500%7D%7D%7B32%7D%3D2.83)
![t=\frac{10-\sqrt{6500}}{32}=-2.2](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B10-%5Csqrt%7B6500%7D%7D%7B32%7D%3D-2.2)
Time cannot be negative .Therefore, negative value of t can not be possible.
Hence, the rock takes 2.83 sec to reach the ground.
If you do in fact mean
(as opposed to one of these being the derivative of
at some point), then integrating twice gives
![f''(x) = -\dfrac1{x^2}](https://tex.z-dn.net/?f=f%27%27%28x%29%20%3D%20-%5Cdfrac1%7Bx%5E2%7D)
![f'(x) = \displaystyle -\int \frac{dx}{x^2} = \frac1x + C_1](https://tex.z-dn.net/?f=f%27%28x%29%20%3D%20%5Cdisplaystyle%20-%5Cint%20%5Cfrac%7Bdx%7D%7Bx%5E2%7D%20%3D%20%5Cfrac1x%20%2B%20C_1)
![f(x) = \displaystyle \int \left(\frac1x + C_1\right) \, dx = \ln|x| + C_1x + C_2](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cdisplaystyle%20%5Cint%20%5Cleft%28%5Cfrac1x%20%2B%20C_1%5Cright%29%20%5C%2C%20dx%20%3D%20%5Cln%7Cx%7C%20%2B%20C_1x%20%2B%20C_2)
From the initial conditions, we find
![f(1) = \ln|1| + C_1 + C_2 = 0 \implies C_1 + C_2 = 0](https://tex.z-dn.net/?f=f%281%29%20%3D%20%5Cln%7C1%7C%20%2B%20C_1%20%2B%20C_2%20%3D%200%20%5Cimplies%20C_1%20%2B%20C_2%20%3D%200)
![f(6) = \ln|6| + 6C_1 + C_2 = 0 \implies 6C_1 + C_2 = -\ln(6)](https://tex.z-dn.net/?f=f%286%29%20%3D%20%5Cln%7C6%7C%20%2B%206C_1%20%2B%20C_2%20%3D%200%20%5Cimplies%206C_1%20%2B%20C_2%20%3D%20-%5Cln%286%29)
Eliminating
, we get
![(C_1 + C_2) - (6C_1 + C_2) = 0 - (-\ln(6))](https://tex.z-dn.net/?f=%28C_1%20%2B%20C_2%29%20-%20%286C_1%20%2B%20C_2%29%20%3D%200%20-%20%28-%5Cln%286%29%29)
![-5C_1 = \ln(6)](https://tex.z-dn.net/?f=-5C_1%20%3D%20%5Cln%286%29)
![C_1 = -\dfrac{\ln(6)}5 = -\ln\left(\sqrt[5]{6}\right) \implies C_2 = \ln\left(\sqrt[5]{6}\right)](https://tex.z-dn.net/?f=C_1%20%3D%20-%5Cdfrac%7B%5Cln%286%29%7D5%20%3D%20-%5Cln%5Cleft%28%5Csqrt%5B5%5D%7B6%7D%5Cright%29%20%5Cimplies%20C_2%20%3D%20%5Cln%5Cleft%28%5Csqrt%5B5%5D%7B6%7D%5Cright%29)
Then
![\boxed{f(x) = \ln|x| - \ln\left(\sqrt[5]{6}\right)\,x + \ln\left(\sqrt[5]{6}\right)}](https://tex.z-dn.net/?f=%5Cboxed%7Bf%28x%29%20%3D%20%5Cln%7Cx%7C%20-%20%5Cln%5Cleft%28%5Csqrt%5B5%5D%7B6%7D%5Cright%29%5C%2Cx%20%2B%20%5Cln%5Cleft%28%5Csqrt%5B5%5D%7B6%7D%5Cright%29%7D)