You serve a volleyball with a mass of 2 kg. The ball leaves your hand with a speed of 30 m/s. The ball has kinetic energy. Calcu
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Answer:
For The value of a = 0 and c = 0 , The given expression equality is true
Step-by-step explanation:
Given expression as :
= a - 2
Or, 3 a² + ac + 2 c - 6 a = ( a - 2 ) × ( 3 a - c )
Or, 3 a² + ac + 2 c - 6 a = 3 a² - ac - 6 a + 2 c
Or, ( 3 a² + ac + 2 c - 6 a ) - ( 3 a² - ac - 6 a + 2 c ) =0
Or, ( 3 a² - 3 a² ) + ( ac + ac ) + ( 2 c - 2 c ) + ( - 6 a + 6 a ) = 0
or, 0 + 2 ac + 0
Or, 2 ac = 0
∴ a =0 and c = 0
Hence For The value of a = 0 and c = 0 , The given expression equality is true . Answer
Answers
b = 2.77 m
A = 43.0°
C = 111.1°
cosine law to find b
b = 2.7708\ m
Find angle A with sine law
![\displaystyle \frac{\sin A}{a} = \frac{\sin B}{b} \\ \\ \sin A = \frac{a \sin B}{b} \\ \\ A = \sin^{-1} \left[ \frac{a \sin B}{b} \right] \\ \\ A = \sin^{-1} \left[ \frac{4.33 \sin 25.9}{2.7708} \right] \\ \\ A = 43.0467020](https://tex.z-dn.net/?f=%5Cdisplaystyle%0A%5Cfrac%7B%5Csin%20A%7D%7Ba%7D%20%3D%20%5Cfrac%7B%5Csin%20B%7D%7Bb%7D%20%5C%5C%20%5C%5C%0A%5Csin%20A%20%3D%20%5Cfrac%7Ba%20%5Csin%20B%7D%7Bb%7D%20%5C%5C%20%5C%5C%0AA%20%3D%20%5Csin%5E%7B-1%7D%20%5Cleft%5B%20%5Cfrac%7Ba%20%5Csin%20B%7D%7Bb%7D%20%20%5Cright%5D%20%5C%5C%20%5C%5C%0AA%20%3D%20%5Csin%5E%7B-1%7D%20%5Cleft%5B%20%5Cfrac%7B4.33%20%5Csin%2025.9%7D%7B2.7708%7D%20%20%5Cright%5D%20%20%5C%5C%20%5C%5C%0AA%20%3D%2043.0467020)
Find C with angles in triangle sum to 180
A + B + C = 180
C = 180 - A - B
C = 180 - 43.0467020 - 25.9
C = 111.1
b = 2.77 m
A = 43.0°
C = 111.1°
cosine law to find b
b = 2.7708\ m
Find angle A with sine law
Find C with angles in triangle sum to 180
A + B + C = 180
C = 180 - A - B
C = 180 - 43.0467020 - 25.9
C = 111.1