Answer:
3, 4, 6, 7, 8, 9 are some examples
Answer:No, it is available in Netflix though <3
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.
Assuming you'd like to have it factored:
Given:
<span>18x^4 + 12x^2y + 2y^2
First take out common factors to all three terms, 2
=2(9x^4+6x^2y+y^2)
Rewrite with x^2 as a group (attempting to factor as perfect square)
=2((3x^2)^2 + 2(3x^2)(y) + y^2)
factor into perfect square
=2(3x^2+y)^2</span>
![{3}^{7} \times {3}^{ - 4}](https://tex.z-dn.net/?f=%20%7B3%7D%5E%7B7%7D%20%20%5Ctimes%20%20%7B3%7D%5E%7B%20-%204%7D%20)
Multiplying to exponents is the same as adding their powers which will be equal to
![{3}^{3}](https://tex.z-dn.net/?f=%20%7B3%7D%5E%7B3%7D%20)
Which is equal to
3 × 3 × 3
9 × 3
27