Answer:
129 - 280 = 51.. So the answer is 51%
∫(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:
An angle is considered a defined term in geometry because defined terms are terms that have a formal definition and can be defined using other geometrical terms.
Angle: Two rays that share the same endpoint, however, the rays take off in different directions. The area in the middle of the two rays is the angle measure.
The answer should be 24 square units.
5 is the hypotenuse and the one we need is the base and the height. They gave us the height, which was 6 in total but 3 for the triangle. But we needed to find the base.
In order to do that, we need to use the Pythagorean Theorem.
a^2+b^2=c^2
3^2+b^2=5^2
9+b^2=25
Subtract 9 from both sides
b^2=16
Then square root both sides.
b=4.
Now that we have the base, you can then find the area of the triangle.
BH/2
4*3/2
12/2
6
So one triangle equals to 6. Then multiply that by 4 to find the area of the rhombus. Which would be 24 square units.
<span>10/3 as a mixed number = 3 1/3
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