Step-by-step explanation:
Given the average of 8 numbers = 56.
Then the Total sum of 8 numbers = 56 * 8
= 448.
Sum of 1st three numbers = 49 + 57 + 72
= 178.
So, the total = 448 - 178
= 270.
Average of the other 5 numbers = 270/5
= 54.
Verification:
178 + 270 = 448/8
= 56.
Answer:
x=16 and y=9
Step-by-step explanation:
2y+30=3y+21[Diagonals of parallelogram are equal]
30-21=3y-2y
y=9
3y=2x-5[Diagonals of parallelogram are equal]
3×9=2x-5
27=2x-5
2x=27+5
x=32/2=16
Prove we are to prove 4(coshx)^3 - 3(coshx) we are asked to prove 4(coshx)^3 - 3(coshx) to be equal to cosh 3x
= 4(e^x+e^(-x))^3/8 - 3(e^x+e^(-x))/2 = e^3x /2 +3e^x /2 + 3e^(-x) /2 + e^(-3x) /2 - 3(e^x+e^(-x))/2 = e^(3x) /2 + e^(-3x) /2 = cosh(3x) = LHS Since y = cosh x satisfies the equation if we replace the "2" with cosh3x, we require cosh 3x = 2 for the solution to work.
i.e. e^(3x)/2 + e^(-3x)/2 = 2
Setting e^(3x) = u, we have u^2 + 1 - 4u = 0
u = (4 + sqrt(12)) / 2 = 2 + sqrt(3), so x = ln((2+sqrt(3))/2) /3, Or u = (4 - sqrt(12)) / 2 = 2 - sqrt(3), so x = ln((2-sqrt(3))/2) /3,
Therefore, y = cosh x = e^(ln((2+sqrt(3))/2) /3) /2 + e^(-ln((2+sqrt(3))/2) /3) /2 = (2+sqrt(3))^(1/3) / 2 + (-2-sqrt(3))^(1/3) to be equ
= 4(e^x+e^(-x))^3/8 - 3(e^x+e^(-x))/2
= e^3x /2 +3e^x /2 + 3e^(-x) /2 + e^(-3x) /2 - 3(e^x+e^(-x))/2
= e^(3x) /2 + e^(-3x) /2
= cosh(3x)
= LHS
<span>Therefore, because y = cosh x satisfies the equation IF we replace the "2" with cosh3x, we require cosh 3x = 2 for the solution to work. </span>
i.e. e^(3x)/2 + e^(-3x)/2 = 2
Setting e^(3x) = u, we have u^2 + 1 - 4u = 0
u = (4 + sqrt(12)) / 2 = 2 + sqrt(3), so x = ln((2+sqrt(3))/2) /3,
Or u = (4 - sqrt(12)) / 2 = 2 - sqrt(3), so x = ln((2-sqrt(3))/2) /3,
Therefore, y = cosh x = e^(ln((2+sqrt(3))/2) /3) /2 + e^(-ln((2+sqrt(3))/2) /3) /2
= (2+sqrt(3))^(1/3) / 2 + (-2-sqrt(3))^(1/3)
Answer:
9x-8 is the simplified form
Answer:
Incomplete question
Complete question: Jaclyn plays singles for South's varsity tennis team. During the match against North, Jaclyn won the sudden death tiebreaker point with a cross-court passing shot. The 57.5-gram ball hit her racket with a northward velocity of 26.7 m/s. Upon impact with her 331-gram racket, the ball rebounded in the exact opposite direction (and along the same general trajectory) with a speed of 29.5 m/s.
a. Determine the pre-collision momentum of the ball.
b. Determine the post-collision momentum of the ball.
c. Determine the momentum change of the ball.
Answer:
A. 1.5353kgm/s
B. 1.6963kgm/s
C. 0.161kgm/s
Step-by-step explanation:
A. The pre-collision momentum of the ball = mass of ball × velocity of ball
Mass of ball = 57.5g = 0.0575kg
Velocity of ball = 26.7m/s
Pre-collision momentum of ball = 0.0575×26.7
= 1.5353kgm/s
B. Post collision momentum of the ball = mass of ball × velocity of ball after impact
Velocity of ball after impact = 29.5m/s
Post collision momentum of ball after impact = 0.0575×29.5
= 1.6963kgm/s
C. Momentum change of ball = momentum after impact - momentum before imlact
= 1.6963kgm/s - 1.5353kgm/s
= 0.161kgm/s
