Answer:
endo 2 17uni
Step-by-step explanation:
5. m∠C = 95°
6. m∠C = 70°
7. The other acute angle in the right triangle = 70°
8. m∠C = 70°
9. m∠C = 60° [equilateral triangle]
10. Measure of the exterior angle at ∠C = 110°
11. m∠B = 70°
12. m∠Z = 70°
<h3>What are Triangles?</h3>
A triangle is a 3-sided polygon with three sides and three angles. The sum of all its interior angles is 180 degrees. Some special triangles are:
- Isosceles triangle: has 2 equal base angles.
- Equilateral triangle: has three equal angles, each measuring 60 degrees.
- Right Triangle: Has one of its angles as 90 degrees, while the other two are acute angles.
5. m∠C = 180 - 50 - 35 [triangle sum theorem]
m∠C = 95°
6. m∠C = 180 - 25 - 85 [triangle sum theorem]
m∠C = 70°
7. The other acute angle in the right triangle = 180 - 90 - 25 [triangle sum theorem]
The other acute angle = 70°
8. m∠C = 180 - 55 - 55 [isosceles triangle]
m∠C = 70°
9. m∠C = 60° [equilateral triangle]
10. Measure of the exterior angle at ∠C = 50 + 60
Measure of the exterior angle at ∠C = 110°
11. m∠B = 115 - 45
m∠B = 70°
12. m∠Z = 180 - 35 - 75
m∠Z = 70°
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Multiply both sides by 2 to cancel out the 1/2 and you get 2a=bh then divide both sides by h to isolate b and you get (2a)/h=b
Answer:
-4 and -2
Step-by-step explanation:
First find all the factors of 8, that would be 1 and 8, -1 and -8, 2 and 4, and -2 and -4.
Then, adding the groups of numbers you have should show you what would equal -6. So 1 + 8 = 9, -1 + (-8) = -9, same vice versa, 2 + 4 = 6, -2 + (-4) = -6. So -2 and -4 would be your two numbers.
∫(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>.
