Let x be the number of rows and y be the number of trees per row.
x* y = 112
and y = 2x - 2
from first equation y = 112/x
so 112/x = 2x - 2
112 = 2x^2 - 2x
2x^2 - 2x - 112 = 0
x^2 - x - 56 = 0
(x - 8)(x + 7) = 0
x = 8 , - 7 (ignore the negative)
so the number of rows = 8
and number of trees per row = 112/8 = 14
Answer number of rows = 8 and number trees i each row = 14.
Answer:
15.7% of students made above an 89.
Step-by-step explanation:
If the data is normally distributed, the standard deviation is 7, and the mean is 82, then about 68.2% of students made between 75 and 89. 13.6% made between 90 and 96, and 2.1% made over 96. 13.6+2.1=15.7%
Given:
u = 44 ft/s, the initial upward velocity of the ball
The height in feet after t seconds is
y = 44t - 16t²
Initial time is t₁ = 2 s.
The height at 2 s is
h₁ = 44*2 - 16*(2²) = 24 ft
(i) When the duration is 0.5 s, the final time is
t₂ = 2 +0.5 = 2.5
The height at 2.5 s is
h₂ = 44*2.5 - 16(2.5²) = 10 ft
The average velocity is
v = (10 - 24 ft)/(0.5 s) = -28 ft/s
Answer: - 28 ft/s
(ii) When the duration is 001s, then t₂ = 2.1 s
h₂ = 44*2.1 - 16(2.1²) = 21.84 ft
The average velocity is
v = (21.84 - 24)/0.1 = -21.6 ft/s
Answer: - 21.6 ft/s
(iii) When the duration is 0.05s, t₂ = 2.05 s
h₂ = 44*2.05 - 16(2.05²) = 22.96
Average velocity is
v = (22.96 - 24)/.05 = - 20.8 ft/s
Answer: - 20.8 ft/s
(iv) When the duration is 0.01 s, t₂ = 2.01 s
h₂ = 44*201 - 16(2.01²) = 23.7984 ft
v = (23.7984 - 24)/0.01 = - 20.16 ft/s
Answer: - 20.16 ft/s
As the duration gets smaller, the computed average velocity approaches the true value at 2 s, because it is the derivative of y with respect to t.