I think the answer to your problem is exponents.
0.2 1/2 because it is closes to two $o yeah.
Angles ∠ACD and ∠CAB are congruent because they are alternate angles. Then the area of the triangle AOB will be 22.53 square cm.
<h3>What is the
area of the right-angle triangle?</h3>
The area of the right-angle triangle is given as
A = 1/2 x B x H
Where B is the base and H is the height of the right triangle.
We know that angles ∠ACD and ∠CAB are congruent because they are alternate angles.
α₁ = 40°
AO = OC = 7.8 cm
Then the area of the triangle will be
Area = 1/2 x 7.8 x 7.8 x tan40°
Area = 22.53 square cm
More about the area of the right-angle triangle link is given below.
brainly.com/question/16653962
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Answer:
The result of the integral is ![-\frac{\cos^3{10x}}{30} + C](https://tex.z-dn.net/?f=-%5Cfrac%7B%5Ccos%5E3%7B10x%7D%7D%7B30%7D%20%2B%20C)
Step-by-step explanation:
We are given the following integral:
![\int \sin{(10x)}\cos^{2}{(10x)} dx](https://tex.z-dn.net/?f=%5Cint%20%5Csin%7B%2810x%29%7D%5Ccos%5E%7B2%7D%7B%2810x%29%7D%20dx)
I am going to solve by substitution, using
, so
. So, we have
![-\frac{1}{10} \int u^2 du](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B10%7D%20%5Cint%20u%5E2%20du)
Which has the following result:
![-\frac{u^3}{30} + C](https://tex.z-dn.net/?f=-%5Cfrac%7Bu%5E3%7D%7B30%7D%20%2B%20C)
Going back to x, the result of the integral is
![-\frac{\cos^3{10x}}{30} + C](https://tex.z-dn.net/?f=-%5Cfrac%7B%5Ccos%5E3%7B10x%7D%7D%7B30%7D%20%2B%20C)
Answer:
A is -8
B 27
x 3
y 6
z 12
step by step explanation: