Suppose a triangle has sides a, b, and c, and that a^2+b^2> c^2. let theta be the measure of the angle opposite the side of length c. which of the following must be true?

check all that apply.

a. theta or delta is an acute angle.
b. the triangle is not a right triangle.
c. cos0> 0
d. the triangle is a right triangle.

Answer with explanation:

It is given that , a triangle having side length , a , b and c follow the Inequality

 →→ a² + b² > c²

→c²-a²-b² <0 or, a²+b²-c²>0 (1)

Law of Cosines

c²=a²+b²-2 a b cos C

→ a²+b²-c²= 2 a b cos C

→ 2 a b cos C >0[Using (1)]

Here, ∠C=theta

So, cos (Theta) > 0

B. The triangle is not a right triangle.

Option B and Option C

hope it

step-by-step explanation:

x^(1/6) · x^(1/6) · x^(1/6)   ·x^(1/6) = x^(4/6)   since we add the exponents

4/6 can be simplified to 2/3

the simplified form is x^(2/3)

it’s supposed to be -4+45=41

step-by-step explanation: