They’re similar due to AAA, each triangle is 180 degrees. since both of the triangles had a degrees of 60 i imputed 67 to the left triangle and they both equal each other
Answer:
Step-by-step explanation:
175 is the answer I don’t feel like working it out
Answer:
![\sqrt[5]{2^4}](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7B2%5E4%7D)
Step-by-step explanation:
Maybe you want 2^(4/5) in radical form.
The denominator of the fractional power is the index of the root. Either the inside or the outside can be raised to the power of the numerator.
![2^{\frac{4}{5}}=\boxed{\sqrt[5]{2^4}=(\sqrt[5]{2})^4}](https://tex.z-dn.net/?f=2%5E%7B%5Cfrac%7B4%7D%7B5%7D%7D%3D%5Cboxed%7B%5Csqrt%5B5%5D%7B2%5E4%7D%3D%28%5Csqrt%5B5%5D%7B2%7D%29%5E4%7D)
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In many cases, it is preferred to keep the power inside the radical symbol.
Answer:
m∠A ≈ 43°
m∠B ≈ 55°
mBC ≈ 20
Step-by-step explanation:
Law of Sines: 
Step 1: Find m∠B

Step 2: Solve for ∠B
29sinB = 24sin82°
sinB = 24sin82°/29
B = sin⁻¹(24sin82°/29)
B = 55.038°
Step 3: Find m∠A
180 - (55.038 + 82)
180 - 137.038
m∠A = 42.962°
Step 4: Find BC

Step 5: Solve for BC
29sin42.962° = BCsin82°
BC = 29sin42.962°/sin82°
BC = 19.9581
W² - v + 1
(-2)² - (-8) + 1
4 + 8 + 1 = 13
B.13 is your answer
(two negatives = one positive)*
(plug in -2 for w, and -8 for v)
(is that w², or do you mean 2w?)
hope this helps