It would be (3) because the measure of angle B is simply 180 degrees minus the measure of angle ACD
Answer:
For 45 guests, Nathan should order:
21.6 (or 22) pieces of Fried Chicken
3.3 pounds of Deli Meat
9.675 pounds of Lasagna
Step-by-step explanation:
The information given for each food item is based on 25 people. If Nathan is going to have 45 guests and plans to only count them as 'half a guest', then we can first take the total number of guests and divide by 2: 45 ÷ 2 = 22.5. Since the original amounts are based on 25 people, we need to find the percentage of food needed for only 22.5 people: 22.5 ÷ 25 = 0.9 x 100 = 90%. Since Nathan will only need 90% of the food totals given, we can find how much he should order by multiplying each food amount by 90% or 0.9:
24 x 0.9 = 21.6 or 22 pieces of Fried Chicken assuming you can't buy partial pieces of chicken
3 2/3 or 11/3 x 0.9 = 3.3 pounds of Deli Meats
10 3/4 or 10.75 x 0.9 = 9.675 pounds of Lasagna
Answer:
It must be a solution for both to be a solution to this system
Im going to go with B.The intersection point (x,y) of the two lines.
Answer:![12.74 ft^2/s](https://tex.z-dn.net/?f=12.74%20ft%5E2%2Fs)
Step-by-step explanation:
Given
Two sides of triangle of sides 5 ft and 7 ft
and angle between them is increasing at a rate of 0.9 radians per second
let
is the angle between them thus
Area of triangle when two sides and angle between them is given
![A=\frac{ab\sin C}{2}](https://tex.z-dn.net/?f=A%3D%5Cfrac%7Bab%5Csin%20C%7D%7B2%7D)
![A=\frac{5\times 7\times \sin \theta }{2}](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B5%5Ctimes%207%5Ctimes%20%5Csin%20%5Ctheta%20%7D%7B2%7D)
Differentiate w.r.t time
![\frac{\mathrm{d} A}{\mathrm{d} t}=\frac{35\cos theta }{2}\times \frac{\mathrm{d} \theta }{\mathrm{d} t}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cmathrm%7Bd%7D%20A%7D%7B%5Cmathrm%7Bd%7D%20t%7D%3D%5Cfrac%7B35%5Ccos%20theta%20%7D%7B2%7D%5Ctimes%20%5Cfrac%7B%5Cmathrm%7Bd%7D%20%5Ctheta%20%7D%7B%5Cmathrm%7Bd%7D%20t%7D)
at ![\theta =\frac{\pi }{5}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%5Cfrac%7B%5Cpi%20%7D%7B5%7D)
![\frac{\mathrm{d} A}{\mathrm{d} t}=\frac{35\times cos(\frac{\pi }{5})}{2}\times 0.9](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cmathrm%7Bd%7D%20A%7D%7B%5Cmathrm%7Bd%7D%20t%7D%3D%5Cfrac%7B35%5Ctimes%20cos%28%5Cfrac%7B%5Cpi%20%7D%7B5%7D%29%7D%7B2%7D%5Ctimes%200.9)
![\frac{\mathrm{d} A}{\mathrm{d} t}=12.74 ft^2/s](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cmathrm%7Bd%7D%20A%7D%7B%5Cmathrm%7Bd%7D%20t%7D%3D12.74%20ft%5E2%2Fs)
Case a: A student can receive any number of awards
Let's count our choices: the first award can go to any of the 20 students. So we have 20 choices. The second awards can also go to any of the 20 students. So we have 20*20 choices for the first two awards. Similarly, we have 20*20*20 choices for the first three awards, and so on.
So, there are
possible ways to give the awards, if a student can receive as many awards as possible.
Case b: A student can receive only one awards
This will be very similar to the previous case, but with a minor restriction: as before, we have 20 choices for the first award, because it can go to any of the 20 students.
But when it comes to the second award, we only have 19 choices, because we can't give it to the student who already won the first award.
Similarly, we can give the third award to one of the 18 remaining students, because we can't give it to the students who already won the first or second award.
So, in the end, we have
ways of awarding the students, if a student can win only one award.