In a triangle HIJ, i = 6.6 inches, j = 6.4 inches and Angle, H=62º. Find the length of h, to the nearest 10th of an inch.

7.6 inches

Step-by-step explanation:

We solve the above question using the Law of Cosines.

The formula is given as:

h² = i² + j² - 2ij Cos H

h = √i² + j² - 2ij Cos H

From the above question:

i = 6.6 inches

j = 6.4 inches

Length h = ?

Angle H=62º.

Hence:

h= √6.6² + 6.4² - 2 × 6.6 × 64 × Cos 62

h = √43.56 + 40.96 - 84.48 × Cos 62

h = 7.64292 inches

Approximately to the nearest tenth, Length h = 7.6 inches

-8

step-by-step explanation:

80x^8−68x^6+16^5−20x^4−4x^3+8x^2

hope i'm right lol