given a triangle with a perimeter of 56in. we know the first side is 3 inches less than twice the length of the second side, and the third side is 10 inches more than half the length of the second side. if the second side length is h, how long is the longest side? write an equation to solve for the variable

Side A = 25 in

Side B = 14 in

Side C = 17 in

Step-by-step explanation:

Perimeter of a triangle = 56 in

Side B = x

Side A = 2x - 3

Side C = 1/2x + 10

Perimeter of a triangle = side A + Side B + side C

56 = (2x -3) + x + (1/2x + 10)

56 = 2x - 3 + x + 1/2x + 10

56 + 3 - 10 = 2x + x + 1/2x

49 = 3 1/2x

Note: 3 1/2x = 3.5x

49 = 3.5x

x = 49/3.5

x = 14 in

Side B = x = 14 in

Side A = 2x - 3

= 2(14) - 3

= 28 - 3

= 25 in

Side C = 1/2x + 10

= 1/2(14) + 10

= 7 + 10

= 17 in

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: