You have to prove that point A is the midpoint of the line segment CB and that will rove that it is equidistant from points c and b
Add -46 and -46 to the final answer
13. you need a total of 14 a week and you have 1
Answer:
Step-by-step explanation:
![\text{Use}\\\\a^\frac{m}{n}=\sqrt[n]{a^m}\\\\a^n\cdot a^m=a^{n+m}\\-----------------\\\\length=\sqrt[3]{81}=\sqrt[3]{3^4}=3^\frac{4}{3}\\\\width=3^\frac{2}{3}\\\\\text{The area of a rectangle:}\ A=width\cdot length.\\\text{Substitute:}\\\\A=3^\frac{4}{3}\cdot3^\frac{2}{3}=3^{\frac{4}{3}+\frac{2}{3}}=3^{\frac{4+2}{3}}=3^\frac{6}{3}=3^2=9](https://tex.z-dn.net/?f=%5Ctext%7BUse%7D%5C%5C%5C%5Ca%5E%5Cfrac%7Bm%7D%7Bn%7D%3D%5Csqrt%5Bn%5D%7Ba%5Em%7D%5C%5C%5C%5Ca%5En%5Ccdot%20a%5Em%3Da%5E%7Bn%2Bm%7D%5C%5C-----------------%5C%5C%5C%5Clength%3D%5Csqrt%5B3%5D%7B81%7D%3D%5Csqrt%5B3%5D%7B3%5E4%7D%3D3%5E%5Cfrac%7B4%7D%7B3%7D%5C%5C%5C%5Cwidth%3D3%5E%5Cfrac%7B2%7D%7B3%7D%5C%5C%5C%5C%5Ctext%7BThe%20area%20of%20a%20rectangle%3A%7D%5C%20A%3Dwidth%5Ccdot%20length.%5C%5C%5Ctext%7BSubstitute%3A%7D%5C%5C%5C%5CA%3D3%5E%5Cfrac%7B4%7D%7B3%7D%5Ccdot3%5E%5Cfrac%7B2%7D%7B3%7D%3D3%5E%7B%5Cfrac%7B4%7D%7B3%7D%2B%5Cfrac%7B2%7D%7B3%7D%7D%3D3%5E%7B%5Cfrac%7B4%2B2%7D%7B3%7D%7D%3D3%5E%5Cfrac%7B6%7D%7B3%7D%3D3%5E2%3D9)
X + 2y = 50
-2x + 3y = 40
5x + 5y + 5z = 300
================
x + 2y = 50 ....multiply by 2
-2x + 3y = 40
--------------------
2x + 4y = 100 (result of multiplying by 2)
-2x + 3y = 40
------------------add
7y = 140
y = 140/7
y = 20 <==
x + 2y = 50
x + 2(20) = 50
x + 40 = 50
x = 50 - 40
x = 10 <==
5x + 5y + 5z = 300
5(10) + 5(20) + 5z = 300
50 + 100 + 5z = 300
150 + 5z = 300
5z = 300 - 150
5z = 150
z = 150/5
z = 30 <==
solution : (10,20,30)...x =10, y = 20 , z = 30
