1) Divide 3/4 by 1/2. Remember how to divide fractions. Switch the problem into a multiplication problem.
<span>3/4 x 2/1 = 3/2 (so you could only have 1 and 1/2 servings with 1/2 pound of chicken in each serving. </span>
<span>2) Set up a little algebra problem. Let x = the multiplier of the original 3/4 pounds of chicken needed to make two full servings with 1/2 pound of chicken in each. </span>
<span>(3/4)x = 2(1/2) (Found this online V(0-0)v)</span>
the perimeter will then just be the sum of the distances of A, B and C, namely AB + BC + CA.
![\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-2})\qquadB(\stackrel{x_2}{0}~,~\stackrel{y_2}{5})\qquad \qquadd = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}\\\\\\AB=\sqrt{[0-(-2)]^2+[5-(-2)]^2}\implies AB=\sqrt{(0+2)^2+(5+2)^2}\\\\\\AB=\sqrt{4+49}\implies \boxed{AB=\sqrt{53}}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\B(\stackrel{x_2}{0}~,~\stackrel{y_2}{5})\qquad C(\stackrel{x_1}{3}~,~\stackrel{y_1}{1})\\\\\\BC=\sqrt{(3-0)^2+(1-5)^2}\implies BC=\sqrt{3^2+(-4)^2}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%5C%5C%5C%5CA%28%5Cstackrel%7Bx_1%7D%7B-2%7D~%2C~%5Cstackrel%7By_1%7D%7B-2%7D%29%5CqquadB%28%5Cstackrel%7Bx_2%7D%7B0%7D~%2C~%5Cstackrel%7By_2%7D%7B5%7D%29%5Cqquad%20%5Cqquadd%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%5C%5C%5C%5C%5C%5CAB%3D%5Csqrt%7B%5B0-%28-2%29%5D%5E2%2B%5B5-%28-2%29%5D%5E2%7D%5Cimplies%20AB%3D%5Csqrt%7B%280%2B2%29%5E2%2B%285%2B2%29%5E2%7D%5C%5C%5C%5C%5C%5CAB%3D%5Csqrt%7B4%2B49%7D%5Cimplies%20%5Cboxed%7BAB%3D%5Csqrt%7B53%7D%7D%5C%5C%5C%5C%5B-0.35em%5D%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5CB%28%5Cstackrel%7Bx_2%7D%7B0%7D~%2C~%5Cstackrel%7By_2%7D%7B5%7D%29%5Cqquad%20C%28%5Cstackrel%7Bx_1%7D%7B3%7D~%2C~%5Cstackrel%7By_1%7D%7B1%7D%29%5C%5C%5C%5C%5C%5CBC%3D%5Csqrt%7B%283-0%29%5E2%2B%281-5%29%5E2%7D%5Cimplies%20BC%3D%5Csqrt%7B3%5E2%2B%28-4%29%5E2%7D)
![\bf BC=\sqrt{9+16}\implies \boxed{BC=5}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\C(\stackrel{x_2}{3}~,~\stackrel{y_2}{1})\qquad A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-2})\\\\\\CA=\sqrt{(-2-3)^2+(-2-1)^2}\implies CA=\sqrt{(-5)^2+(-3)^2}\\\\\\CA=\sqrt{25+9}\implies \boxed{CA=\sqrt{34}}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\~\hfill \stackrel{AB+BC+CA}{\approx 18.11}~\hfill](https://tex.z-dn.net/?f=%5Cbf%20BC%3D%5Csqrt%7B9%2B16%7D%5Cimplies%20%5Cboxed%7BBC%3D5%7D%5C%5C%5C%5C%5B-0.35em%5D%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5CC%28%5Cstackrel%7Bx_2%7D%7B3%7D~%2C~%5Cstackrel%7By_2%7D%7B1%7D%29%5Cqquad%20A%28%5Cstackrel%7Bx_1%7D%7B-2%7D~%2C~%5Cstackrel%7By_1%7D%7B-2%7D%29%5C%5C%5C%5C%5C%5CCA%3D%5Csqrt%7B%28-2-3%29%5E2%2B%28-2-1%29%5E2%7D%5Cimplies%20CA%3D%5Csqrt%7B%28-5%29%5E2%2B%28-3%29%5E2%7D%5C%5C%5C%5C%5C%5CCA%3D%5Csqrt%7B25%2B9%7D%5Cimplies%20%5Cboxed%7BCA%3D%5Csqrt%7B34%7D%7D%5C%5C%5C%5C%5B-0.35em%5D%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C~%5Chfill%20%5Cstackrel%7BAB%2BBC%2BCA%7D%7B%5Capprox%2018.11%7D~%5Chfill)
Answer:
6
Step-by-step explanation:
All of them. (6)
Let x and y be the rate per hour for Mechanic 1 and 2 respectively.
Then,
x+y=205 ---------- (1)
Additionally,
10x+15y = 2500 ------ (2)
Solving equations 1 and 2;
Multiply (1) by 10 and subtract two from the result,
10x+10y = 2050
10x +15 y =2500
----------------------
-5 y = --450 => y = 450/5 = $90
Then,
x= 205-90 = $115
Therefore,
Mechanic 1 charged $115 per hour
Mechanic 2 charged $90 per hour
