Answer: see proof below
<u>Step-by-step explanation:</u>
Given: A + B + C = π → A = π - (B + C)
→ B = π - (A + C)
→ C = π - (A + B)
Use Sum to Product Identity: sin A - sin B = 2 cos [(A + B)/2] · sin [(A - B)/2]
Use the following Cofunction Identity: cos (π/2 - A) = sin A
<u>Proof LHS → RHS:</u>
LHS: sin A - sin B + sin C
= (sin A - sin B) + sin C
![\text{Factor:}\qquad 2\sin \bigg(\dfrac{C}{2}\bigg)\bigg[ \sin \bigg(\dfrac{A-B}{2}\bigg)+\cos \bigg(\dfrac{C}{2}\bigg)\bigg]](https://tex.z-dn.net/?f=%5Ctext%7BFactor%3A%7D%5Cqquad%202%5Csin%20%5Cbigg%28%5Cdfrac%7BC%7D%7B2%7D%5Cbigg%29%5Cbigg%5B%20%5Csin%20%5Cbigg%28%5Cdfrac%7BA-B%7D%7B2%7D%5Cbigg%29%2B%5Ccos%20%5Cbigg%28%5Cdfrac%7BC%7D%7B2%7D%5Cbigg%29%5Cbigg%5D)
![\text{Given:}\qquad 2\sin \bigg(\dfrac{C}{2}\bigg)\bigg[ \sin \bigg(\dfrac{A-B}{2}\bigg)+\cos \bigg(\dfrac{\pi -(A+B)}{2}\bigg)\bigg]\\\\\\.\qquad \qquad =2\sin \bigg(\dfrac{C}{2}\bigg)\bigg[ \sin \bigg(\dfrac{A-B}{2}\bigg)+\cos \bigg(\dfrac{\pi}{2} -\dfrac{(A+B)}{2}\bigg)\bigg]](https://tex.z-dn.net/?f=%5Ctext%7BGiven%3A%7D%5Cqquad%202%5Csin%20%5Cbigg%28%5Cdfrac%7BC%7D%7B2%7D%5Cbigg%29%5Cbigg%5B%20%5Csin%20%5Cbigg%28%5Cdfrac%7BA-B%7D%7B2%7D%5Cbigg%29%2B%5Ccos%20%5Cbigg%28%5Cdfrac%7B%5Cpi%20-%28A%2BB%29%7D%7B2%7D%5Cbigg%29%5Cbigg%5D%5C%5C%5C%5C%5C%5C.%5Cqquad%20%5Cqquad%20%3D2%5Csin%20%5Cbigg%28%5Cdfrac%7BC%7D%7B2%7D%5Cbigg%29%5Cbigg%5B%20%5Csin%20%5Cbigg%28%5Cdfrac%7BA-B%7D%7B2%7D%5Cbigg%29%2B%5Ccos%20%5Cbigg%28%5Cdfrac%7B%5Cpi%7D%7B2%7D%20-%5Cdfrac%7B%28A%2BB%29%7D%7B2%7D%5Cbigg%29%5Cbigg%5D)
![\text{Cofunction:}\qquad 2\sin \bigg(\dfrac{C}{2}\bigg)\bigg[ \sin \bigg(\dfrac{A-B}{2}\bigg)+\sin \bigg(\dfrac{A+B}{2}\bigg)\bigg]](https://tex.z-dn.net/?f=%5Ctext%7BCofunction%3A%7D%5Cqquad%202%5Csin%20%5Cbigg%28%5Cdfrac%7BC%7D%7B2%7D%5Cbigg%29%5Cbigg%5B%20%5Csin%20%5Cbigg%28%5Cdfrac%7BA-B%7D%7B2%7D%5Cbigg%29%2B%5Csin%20%5Cbigg%28%5Cdfrac%7BA%2BB%7D%7B2%7D%5Cbigg%29%5Cbigg%5D)
![\text{Sum to Product:}\qquad 2\sin \bigg(\dfrac{C}{2}\bigg)\bigg[ 2\sin \bigg(\dfrac{A}{2}\bigg)\cdot \cos \bigg(\dfrac{B}{2}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad \qquad =4\sin \bigg(\dfrac{A}{2}\bigg)\cdot \cos \bigg(\dfrac{B}{2}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)](https://tex.z-dn.net/?f=%5Ctext%7BSum%20to%20Product%3A%7D%5Cqquad%202%5Csin%20%5Cbigg%28%5Cdfrac%7BC%7D%7B2%7D%5Cbigg%29%5Cbigg%5B%202%5Csin%20%5Cbigg%28%5Cdfrac%7BA%7D%7B2%7D%5Cbigg%29%5Ccdot%20%5Ccos%20%5Cbigg%28%5Cdfrac%7BB%7D%7B2%7D%5Cbigg%29%5Cbigg%5D%5C%5C%5C%5C%5C%5C.%5Cqquad%20%5Cqquad%20%5Cqquad%20%5Cqquad%20%3D4%5Csin%20%5Cbigg%28%5Cdfrac%7BA%7D%7B2%7D%5Cbigg%29%5Ccdot%20%5Ccos%20%5Cbigg%28%5Cdfrac%7BB%7D%7B2%7D%5Cbigg%29%5Ccdot%20%5Csin%20%5Cbigg%28%5Cdfrac%7BC%7D%7B2%7D%5Cbigg%29)
You can estimate 3.9 times 5.3 to 4 times 5 which is 20
so if the answer is around 20, you know that you will have exactly 2 digits before the decimal place
the answer is 54 9/14
because Divide using long division. The whole number portion will be the number of times the denominator of the original fraction divides evenly into the numerator of the original fraction, and the fraction portion of the mixed number will be the remainder of the original fraction division over the denominator of the original fraction.
Answer:
2290 trees
Step-by-step explanation:
Given data
Let the total number of trees be x
62% of the total trees x is 1420 trees
mathematically
62/100x= 1420
0.62x= 1420
divide both sides by 0.62
x= 1420/0.62
x= 2290.32
To the nearest whole tree= 2290 trees
Answer:
3(7a - 9)
Step-by-step explanation:
21a–27
3*7*a - 3*9
Factor out 3
3(7a - 9)
