3x+45 most u can simplify it
You need to use a ratio of height (H) to shadow length (L) to solve the first problem. It's basically a use of similar triangles, with two perpendicular sides, and with the shadow making the same angle with the vertical.
6 ft = 72 ins, so that rH/L = 72/16 = 9/2 for the player.
So the bleachers are 9/2 x 6 ft = 27 ft.
For the second problem, 9 ft = 108 in, so that the ratio of the actual linear dimensions to the plan's linear dimensions are 9ft/(1/2in) = 2 x 108 = 216.
So the stage will have dimensions 216 times larger than 1.75" by 3".
That would be 31ft 6ins x 54ft.
Live long and prosper.
Answer:
See Below (Isolate the variables)
Step-by-step explanation:
13) d = 4AM solve for M
Divide both sides by 4A
14) 
= f+h solve for A
Multiply both sides by 6
A = 6( f+h)
15) h - m = qb solve for H
Add m to both sides
h - m = qb
+ m +m
h = qb + m
