Let
be the usual speed, and
be the usual time.
Following the equation
, where s is the space, v is the velocity and t is time, we know that usually we have
![200=vt](https://tex.z-dn.net/?f=200%3Dvt)
But this time we were 10mph faster and it took one hour less, so we have
![200=(v+10)(t-1)](https://tex.z-dn.net/?f=200%3D%28v%2B10%29%28t-1%29)
Since both right hand sides equal 200, they must equal each other:
![vt=(v+10)(t-1)=vt+10t-v-10 \iff 10t-v-10=0 \iff v=10t-10](https://tex.z-dn.net/?f=vt%3D%28v%2B10%29%28t-1%29%3Dvt%2B10t-v-10%20%5Ciff%2010t-v-10%3D0%20%5Ciff%20v%3D10t-10)
Plug this value in the first equation and we have
![200=vt=(10t-10)t \iff 10t^2-10t-200=0](https://tex.z-dn.net/?f=200%3Dvt%3D%2810t-10%29t%20%5Ciff%2010t%5E2-10t-200%3D0)
This equation has solutions
. We can only accept positive solutions, so we have t=5. We finally deduce, again from the first equation,
![200=5v \iff v=\dfrac{200}{5}=40](https://tex.z-dn.net/?f=200%3D5v%20%5Ciff%20v%3D%5Cdfrac%7B200%7D%7B5%7D%3D40)
Answer:
Maybe is 0,999999999999999....
Step-by-step explanation:
Answer:
5
Step-by-step explanation:
Slope formula=
![\frac{y2-x2}{y1-x1}](https://tex.z-dn.net/?f=%5Cfrac%7By2-x2%7D%7By1-x1%7D)
so the slope is
![\frac{-4-16}{2-6}](https://tex.z-dn.net/?f=%5Cfrac%7B-4-16%7D%7B2-6%7D)
or
![\frac{-20}{-4}](https://tex.z-dn.net/?f=%5Cfrac%7B-20%7D%7B-4%7D)
or![5](https://tex.z-dn.net/?f=5)
so the slope is 5
Hope this helps ;D