Answer:
how to set up problem: (7x-36)+(5x+12)=180
Step-by-step explanation:
The approximate stopping distance d(in feet) is given by the formula :
...(1)
Where
v is the speed of the car in mph
We need to find the speed of the car when the distance is 200 feet
Put d = 200 in equation (1)
![0.05v^2+2.2v=200\\\\0.05v^2+2.2v-200=0](https://tex.z-dn.net/?f=0.05v%5E2%2B2.2v%3D200%5C%5C%5C%5C0.05v%5E2%2B2.2v-200%3D0)
It is a quadratic equation. Divide the above equation by 0.05.
![v^2+44v-4000=0](https://tex.z-dn.net/?f=v%5E2%2B44v-4000%3D0)
The solution of the above equation is given by :
![v=\dfrac{-44\pm \sqrt{44^2- 4(1)(-4000)} }{2(1)}\\\\v=\dfrac{-44+ \sqrt{44^2- 4(1)(-4000)} }{2(1)},\dfrac{-44- \sqrt{44^2- 4(1)(-4000)} }{2(1)}\\\\v=44.96\ ft/h, -88.96\ ft/h](https://tex.z-dn.net/?f=v%3D%5Cdfrac%7B-44%5Cpm%20%5Csqrt%7B44%5E2-%204%281%29%28-4000%29%7D%20%7D%7B2%281%29%7D%5C%5C%5C%5Cv%3D%5Cdfrac%7B-44%2B%20%5Csqrt%7B44%5E2-%204%281%29%28-4000%29%7D%20%7D%7B2%281%29%7D%2C%5Cdfrac%7B-44-%20%5Csqrt%7B44%5E2-%204%281%29%28-4000%29%7D%20%7D%7B2%281%29%7D%5C%5C%5C%5Cv%3D44.96%5C%20ft%2Fh%2C%20-88.96%5C%20ft%2Fh)
Since, 1 mph = 5280 ft/hour
44.96 ft/h = 0.00851 mph
-88.96 = -0.0168 mph
Hence, this is the required siolution.
You would know because 1/4 = .25 and are interchangeablemaking the equations the same <span />
Answer:
Sorry its late
Step-by-step explanation:
A dot plot titled Distance from School in Blocks going from 1 to 6. 1 has 6 dots, 2 has 5 dots, 3 has 4 dots, 4 has 2 dots, 5 has 3 dots, and 6 has 2 dots.
Wynn needs to find the center of the data set shown on the dot plot.
The dot plot has
✔ 22
dots.
The dot plot has
✔ 11
dots to the left of the
center and
✔ 11
dots to the right of the center.
The center of the data set is
✔ between 2 and 3
.
None of the above. Three EVEN integers will never make a sum of 45, which is an ODD number.