Answer: see proof below
<u>Step-by-step explanation:</u>
Given: A + B + C = π → A + B = π - C
→ B + C = π - A
→ C + A = π - B
→ C = π - (B + C)
Use Sum to Product Identity: cos A + cos B = 2 cos [(A + B)/2] · cos [(A - B)/2]
Use the Sum/Difference Identity: cos (A - B) = cos A · cos B + sin A · sin B
Use the Double Angle Identity: sin 2A = 2 sin A · cos A
Use the Cofunction Identity: cos (π/2 - A) = sin A
<u>Proof LHS → Middle:</u>
![\text{Factor:}\quad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\bigg[ \cos \bigg(\dfrac{A-B}{4}\bigg)+\sin \bigg(\dfrac{A+B}{4}\bigg)\bigg]](https://tex.z-dn.net/?f=%5Ctext%7BFactor%3A%7D%5Cquad%20%20%3D2%5Ccos%20%5Cbigg%28%5Cdfrac%7BA%2BB%7D%7B4%7D%5Cbigg%29%5Cbigg%5B%20%5Ccos%20%5Cbigg%28%5Cdfrac%7BA-B%7D%7B4%7D%5Cbigg%29%2B%5Csin%20%5Cbigg%28%5Cdfrac%7BA%2BB%7D%7B4%7D%5Cbigg%29%5Cbigg%5D)
![\text{Cofunction:}\quad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\bigg[ \cos \bigg(\dfrac{A-B}{4}\bigg)+\cos \bigg(\dfrac{\pi}{2}-\dfrac{A+B}{4}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\cos \bigg(\dfrac{2\pi-(A+B)}{4}\bigg)\bigg]](https://tex.z-dn.net/?f=%5Ctext%7BCofunction%3A%7D%5Cquad%20%20%3D2%5Ccos%20%5Cbigg%28%5Cdfrac%7BA%2BB%7D%7B4%7D%5Cbigg%29%5Cbigg%5B%20%5Ccos%20%5Cbigg%28%5Cdfrac%7BA-B%7D%7B4%7D%5Cbigg%29%2B%5Ccos%20%5Cbigg%28%5Cdfrac%7B%5Cpi%7D%7B2%7D-%5Cdfrac%7BA%2BB%7D%7B4%7D%5Cbigg%29%5Cbigg%5D%5C%5C%5C%5C%5C%5C.%5Cqquad%20%5Cqquad%20%5Cqquad%20%3D2%5Ccos%20%5Cbigg%28%5Cdfrac%7BA%2BB%7D%7B4%7D%5Cbigg%29%5Ccdot%20%5Ccos%20%5Cbigg%28%5Cdfrac%7BA-B%7D%7B4%7D%5Cbigg%29%2B%5Ccos%20%5Cbigg%28%5Cdfrac%7B2%5Cpi-%28A%2BB%29%7D%7B4%7D%5Cbigg%29%5Cbigg%5D)
![\text{Sum to Product:}\ 2\cos \bigg(\dfrac{A+B}{4}\bigg)\bigg[2 \cos \bigg(\dfrac{2\pi-2B}{2\cdot 4}\bigg)\cdot \cos \bigg(\dfrac{2A-2\pi}{2\cdot 4}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =4\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi-B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi -A}{4}\bigg)](https://tex.z-dn.net/?f=%5Ctext%7BSum%20to%20Product%3A%7D%5C%202%5Ccos%20%5Cbigg%28%5Cdfrac%7BA%2BB%7D%7B4%7D%5Cbigg%29%5Cbigg%5B2%20%5Ccos%20%5Cbigg%28%5Cdfrac%7B2%5Cpi-2B%7D%7B2%5Ccdot%204%7D%5Cbigg%29%5Ccdot%20%5Ccos%20%5Cbigg%28%5Cdfrac%7B2A-2%5Cpi%7D%7B2%5Ccdot%204%7D%5Cbigg%29%5Cbigg%5D%5C%5C%5C%5C%5C%5C.%5Cqquad%20%5Cqquad%20%5Cqquad%20%3D4%5Ccos%20%5Cbigg%28%5Cdfrac%7BA%2BB%7D%7B4%7D%5Cbigg%29%5Ccdot%20%5Ccos%20%5Cbigg%28%5Cdfrac%7B%5Cpi-B%7D%7B4%7D%5Cbigg%29%5Ccdot%20%5Ccos%20%5Cbigg%28%5Cdfrac%7B%5Cpi%20-A%7D%7B4%7D%5Cbigg%29)
LHS = Middle 
<u>Proof Middle → RHS:</u>
Middle = RHS
Answer:
there is 22 males in the class
Step-by-step explanation:
0.09 Is The Answer! Hope I Helped!
-1 = z/3 - 7
Emily's Work:
-1 = z/3 - 7
-1(3)=z/3(3)-7
-3 = z - 7
-3 -7 = z - 7
z = -10
Right away without having to start at the beginning, I see that she has made a mistake at the 4th-5th line of work.
Instead of adding 7 to both sides, she subtracts it and still leaves it in and doesn't fully finish the equation, as it would really be z-7 = -10
In order to solve it, follow these steps:
Rewrite the initial equation as follows:
x/3 - 7 = -1
Multiply both sides by 3
x/3(3)-7(3)=-1(3)
x - 21 = -3
Add 21 to both sides
x = 18
20% of 65 = 65 * 1/5, which is 13. so 20% of 65 is $13. so a discount of 20% = 65-13 which is $52
