<span>The absolute value of an integer is the numerical valuewithout regard to whether the sign is negative or positive. On a number line it is the distance between the number and zero. Absolute value is also known as magnitude.</span>
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)
Answer:
If you meant [80 + (8*4) ]/2, then [80 + (8*4) ]/2=56
or if you meant 80 + [ 8*4 / 2], then 80 + [ 8*4 / 2] = 96
Step-by-step explanation:
Evalute the expression. 80 + (8x4) divided by 2
so do you mean
the whole expression 80 + (8x4) is over 2 ?
If so then:
evaluate 8x4 = 8 * 4 = 32.
so we have 80 + (8x4) is over 2 = (80 + 32)/2 = 112/2 = 56
If you meant 80 + [ 8*4 / 2]
then 80 + [ 8*4 / 2] = 80 + 8*2 = 80 + 16 = 96
Answer:
333
Step-by-step explanation:
Answer:
Step-by-step explanation:
(a + b + c)³ = a³ + b³ + c³ + 3a²b + 3a²c + 3ab² + 3cb² +3 ac² + 3bc² + 6abc
a = 5a ; b =y ; c = z
(5x + y + z)(5z + y + z )(5z + y +z) = (5x + y +z)³
= (5x)³ + y³ +z³ + 3(5x)²y + 3(5x)²z + 3(5x)*y² + 3*z*y² + 3*5x*z² + 3*y*z² + 6*5x*y*z
= 125x³ + y³ +z³ + 3*25x²y + 3*25x²*z + 15xy² + 3zy² + 15xz² + 3yz² + 35xyz
= 125x³ + y³ + z³ + 75x²y + 75x²z + 15xy² + 3zy² + 15xz² + 3yz² + 35xyz
