The area of the sector is 86.81π mm²
<u>Explanation:</u>
Given:
Radius of the sector, 25mm
Angle, α = 50°
Area of the sector, A = ?
We know:
Area of the sector = ![\frac{\alpha }{360} X \pi r^2](https://tex.z-dn.net/?f=%5Cfrac%7B%5Calpha%20%7D%7B360%7D%20X%20%5Cpi%20r%5E2)
On substituting the value, we get:
![A = \frac{50}{360} X \pi X (25)^2\\\\A = 86.81 \pi mm^2](https://tex.z-dn.net/?f=A%20%3D%20%5Cfrac%7B50%7D%7B360%7D%20X%20%5Cpi%20X%20%2825%29%5E2%5C%5C%5C%5CA%20%3D%2086.81%20%5Cpi%20mm%5E2)
Therefore, the area of the sector is 86.81π mm²
Given that the height of the disc is modeled by the function:
h=-16t^2+20t+6
a] The maximum height will be as follows:
At maximum height:
h'(t)=0
from h(t)
h'(t)=-32t+20=0
thus
t=20/32
t=5/8 sec
thus the maximum height will be:
h(5/8)=-16(5/8)^2+20(5/8)+6
=12.25 ft
b] <span>How long will it take the disc to reach the maximum height?
</span>time taken to reach maximum height will be:
from h(t)
h'(t)=-32t+20=0
thus
t=20/32
t=5/8 sec
thus time taken to reach maximum height is t=5/8 sec
c]<span>How long does it take for the disc to be caught 3 feet off the ground?
</span>h(t)=-16t^2+20t+6
but
h(t)=3
thus
3=-16t^2+20t+6
solving for t we get:
0=-16t^2+20t+3
factoring the above we get:
t=5/8-/+√37/8
t=-1.5256 or 2.776
since there is no negative time we pick t=2.776
Hence time taken for the disc to be caught 3 ft from the ground is 2.776 ft
Answer:
Step-by-step explanation:
2√5
Answer:
![(x,y) = (1,-12)](https://tex.z-dn.net/?f=%28x%2Cy%29%20%3D%20%281%2C-12%29)
Step-by-step explanation:
Given
![y = 6x - 18](https://tex.z-dn.net/?f=y%20%3D%206x%20-%2018)
Required
Find a solution
To do this, we assume a value for x, then solve for y.
Let x = 1
Substitute 1 for x in ![y = 6x - 18](https://tex.z-dn.net/?f=y%20%3D%206x%20-%2018)
![y = 6*1 - 18](https://tex.z-dn.net/?f=y%20%3D%206%2A1%20-%2018)
![y = 6 - 18](https://tex.z-dn.net/?f=y%20%3D%206%20-%2018)
![y= -12](https://tex.z-dn.net/?f=y%3D%20-12)
Hence, a solution is:
![(x,y) = (1,-12)](https://tex.z-dn.net/?f=%28x%2Cy%29%20%3D%20%281%2C-12%29)