Answer:
60 chickens
Step-by-step explanation:
We assume the 140 refers to the number animals (heads, noses, bodies, whatever). If all of those were chickens, there would be 280 legs. There are 160 legs more than that.
Since each pig that replaces a chicken adds 2 legs, there must be 160/2 = 80 pigs. That leaves 140 -80 = 60 chickens.
∫(t = 2 to 3) t^3 dt
= (1/4)t^4 {for t = 2 to 3}
= 65/4.
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∫(t = 2 to 3) t √(t - 2) dt
= ∫(u = 0 to 1) (u + 2) √u du, letting u = t - 2
= ∫(u = 0 to 1) (u^(3/2) + 2u^(1/2)) du
= [(2/5) u^(5/2) + (4/3) u^(3/2)] {for u = 0 to 1}
= 26/15.
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For the k-entry, use integration by parts with
u = t, dv = sin(πt) dt
du = 1 dt, v = (-1/π) cos(πt).
So, ∫(t = 2 to 3) t sin(πt) dt
= (-1/π) t cos(πt) {for t = 2 to 3} - ∫(t = 2 to 3) (-1/π) cos(πt) dt
= (-1/π) (3 * -1 - 2 * 1) + [(1/π^2) sin(πt) {for t = 2 to 3}]
= 5/π + 0
= 5/π.
Therefore,
∫(t = 2 to 3) <t^3, t√(t - 2), t sin(πt)> dt = <65/4, 26/15, 5/π>.
Answer:
B
Step-by-step explanation:
x^3 + 2x^2 - 9x - 18 = x^2(x + 2) - 9(x + 2) = (x + 2)*(x^2 - 9)
(x + 2)/(x + 2)(x^2 - 9) = 1 / (x^2 - 9)
==============
1/(x^2 - 9) * (x^2 - 9)/(3x + 1) = 1/(3x + 1)
Yes B is correct.
Answer:
31.76 ft and 58.64 ft
Step-by-step explanation:
The radius measures between 13 feet and 24 feet.
The wheel is able to turn 7π/9 radians before getting stuck.
We need to find the range of distances that the wheel could spin before getting stuck. That is, the length of arc.
Length of an arc is given as:
where θ = central angle = 7π/9 radians
r = radius of the circle
Therefore, for 13 feet:
For 24 feet:
The wheel could spin between 31.76 ft and 58.64 ft before getting stuck.
Answer:
2.22
Step-by-step explanation:
