∫(t = 2 to 3) t^3 dt
= (1/4)t^4 {for t = 2 to 3}
= 65/4.
----
∫(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.
----
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:
-20
Step-by-step explanation:

According to PEMDAS, we need to do the math in the parentheses first.

Applying PEMDAS to the inside of the parentheses, then we have to divide -6 by -1, where we would get 6.

Adding a negative is the same as subtracting, so adding negative 10 is the same as subtracting 10.

Subtracting, we get 6 - 10 = -4 and we're left with:

Multiplying, <u>our final answer is -20.</u>
Hope this helps!
Answer:
Step-by-step explanation:
Put these numbers into their prime factors.
18: 2 * 3 * 3
27: 3 * 3 * 3
12: 2 * 2 * 3
The LCM must have
two 2s
Three 3s
That's it
2*2 * 3 * 3 * 3
LCM = 108
648 might be a multiple of the three, but it's not the smallest one.
108 = 27 * 4
108 = 18 * 6
108 = 12 * 9
Answer:
20 red blocks
Step-by-step explanation:
Lee blocks--20 + 36= 56
Dan blocks 13 + 23 =36
56 - 36=20