∫(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>.
The answer is D-16
16/4= 4. 4x3=12. 12-2=10
Answer:
Step-by-step explanation:
You are to assume in both problems that the two triangles are similar. That is a very dangerous assumption -- especially in later math classes. But in this case, there is no other way to do the problem. The two sets of triangles look like they are proportional. So set up two ratios that are = to each other
On the left
long side small triange / long side large triangle = base small triangle / base large triangle
On the right
hypotenuse/longest leg = hypotenuse / longest leg.
Problem A
Set up the similar Proportion
5/40 = x/20 Cross Multiply
40x = 20*5
40x = 100 Divide by 40
x = 100/40
x = 2.5
Problem B
Again set up the similar triangle proportion
15/10 = x/2 Multiply by 2
15*2/10 = x
3 = x
Answer:
0.06 and 6/100
Step-by-step explanation:
Answer:
0
x = 2y/0 = 0
x = 2y/2 = y
x = 3y/4
HOPE THIS HELPS YOU
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