Use the Product Rule: x^ax^b = x^a + b
6x^3 + 1y^2 + 6
Simplify 3 + 1 to 4
6x^4y^2 + 6
Simplify 2 + 6 to 8
<u>= C) 6x^4 y^8</u>
Answer:
(13,-10)
(Sorry for no step by step)
Refer to the diagram shown below.
The directrix is y = -4 and the focus is (-2, -2).
Therefore the vertex is at (-2, -3).
Consider an arbitrary point (x,y) on the parabola.
The square of distance from the focus to the point is
(x+2)² + (y+2)²
The square of the distance from the point to the directrix is
(y+4)²
Therefore
(y+4)² = (y+2)² + (x+2)²
y² + 8y + 16 = y² + 4y + 4 + (x+2)²
4y = (x+2)² - 12
y = (1/4)(x+2)² - 3
Answer:
The air force plane travelled for a total of 6.9 hours.
<u>SOLUTION:
</u>
Given, An Air Force plane left Nairobi and flew west at an average speed of 159 mph. A cargo plane left sometime later flying in the same direction at an average speed of 207 mph. After flying for 5.3 hours the cargo plane caught up with the Air Force plane.
We have to find the number of hours the Air Force plane flew before the cargo plane caught up.
Now, we know that, ![\text { distance travelled }=\text { speed } \times \text { time }](https://tex.z-dn.net/?f=%5Ctext%20%7B%20distance%20travelled%20%7D%3D%5Ctext%20%7B%20speed%20%7D%20%5Ctimes%20%5Ctext%20%7B%20time%20%7D)
For air force plane ![\rightarrow \text { distance travelled }=159 \mathrm{x} \text { time taken by air force plane }](https://tex.z-dn.net/?f=%5Crightarrow%20%5Ctext%20%7B%20distance%20travelled%20%7D%3D159%20%5Cmathrm%7Bx%7D%20%5Ctext%20%7B%20time%20taken%20by%20air%20force%20plane%20%7D)
![\text { For cargo plane } \rightarrow \text { distance travelled }=207 \times 5.3 \text { hours }](https://tex.z-dn.net/?f=%5Ctext%20%7B%20For%20cargo%20plane%20%7D%20%5Crightarrow%20%5Ctext%20%7B%20distance%20travelled%20%7D%3D207%20%5Ctimes%205.3%20%5Ctext%20%7B%20hours%20%7D)
Now, when cargo plane caught the air force plane the distance travelled by the planes will be equal.
So, ![159 \times \text { time taken by air force plane }=207 \times 5.3](https://tex.z-dn.net/?f=159%20%5Ctimes%20%5Ctext%20%7B%20time%20taken%20by%20air%20force%20plane%20%7D%3D207%20%5Ctimes%205.3)
![\Rightarrow =\frac{207 \times 5.3}{159}](https://tex.z-dn.net/?f=%5CRightarrow%20%3D%5Cfrac%7B207%20%5Ctimes%205.3%7D%7B159%7D)
![\Rightarrow =\frac{1097.1}{159}](https://tex.z-dn.net/?f=%5CRightarrow%20%3D%5Cfrac%7B1097.1%7D%7B159%7D)
Time taken by air force plane = 6.9 hrs
Can you help me with my posted questions