Answer: (5,2)
Step-by-step explanation:
Given equations: ![2x-2y=6......(1)\\4x+4y=28.......(2)](https://tex.z-dn.net/?f=2x-2y%3D6......%281%29%5C%5C4x%2B4y%3D28.......%282%29)
Divide 2 on both sides of equation (1) and 4 on the both sides of equation (2), we get
![x-y=3......(3)\\x+y=7.......(4)](https://tex.z-dn.net/?f=x-y%3D3......%283%29%5C%5Cx%2By%3D7.......%284%29)
Adding equation (3) and (4), we get
![2x=10\\\Rightarrow\ x=5](https://tex.z-dn.net/?f=2x%3D10%5C%5C%5CRightarrow%5C%20x%3D5)
Substitute x=5 in equation (4), we get
![y=7-5=2](https://tex.z-dn.net/?f=y%3D7-5%3D2)
Hence, the solution of the both equations is (5,2).
if we take 64 to be the 100%, how much is 6¼% off of it?
![\bf \begin{array}{ccll} amount&\%\\ \cline{1-2} 64&100\\ x&6\frac{1}{4} \end{array}\implies \cfrac{64}{x}=\cfrac{100}{6\frac{1}{4}}\implies \cfrac{64}{x}=\cfrac{\frac{100}{1}}{\frac{25}{4}}\implies \cfrac{64}{x}=\cfrac{100}{1}\cdot \cfrac{4}{25} \\\\\\ \cfrac{64}{x}=16\implies 64=16x\implies \cfrac{64}{16}=x\implies 4=x \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{it had}}{64}-\stackrel{\textit{leakage}}{4}\implies \stackrel{\textit{remaining}}{60}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Bccll%7D%20amount%26%5C%25%5C%5C%20%5Ccline%7B1-2%7D%2064%26100%5C%5C%20x%266%5Cfrac%7B1%7D%7B4%7D%20%5Cend%7Barray%7D%5Cimplies%20%5Ccfrac%7B64%7D%7Bx%7D%3D%5Ccfrac%7B100%7D%7B6%5Cfrac%7B1%7D%7B4%7D%7D%5Cimplies%20%5Ccfrac%7B64%7D%7Bx%7D%3D%5Ccfrac%7B%5Cfrac%7B100%7D%7B1%7D%7D%7B%5Cfrac%7B25%7D%7B4%7D%7D%5Cimplies%20%5Ccfrac%7B64%7D%7Bx%7D%3D%5Ccfrac%7B100%7D%7B1%7D%5Ccdot%20%5Ccfrac%7B4%7D%7B25%7D%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B64%7D%7Bx%7D%3D16%5Cimplies%2064%3D16x%5Cimplies%20%5Ccfrac%7B64%7D%7B16%7D%3Dx%5Cimplies%204%3Dx%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bit%20had%7D%7D%7B64%7D-%5Cstackrel%7B%5Ctextit%7Bleakage%7D%7D%7B4%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bremaining%7D%7D%7B60%7D)
Answer:
i think r would be the same as q
Step-by-step explanation:
I believe it's an isosceles triangle, if there were a line going down the center, each half should be symmetrical, so q and r are the same angle.
The area of the house is the amount of space on the house.
- The length of the addition is x + 20
- The area of the original house is
![x^2 + 30x + 200](https://tex.z-dn.net/?f=x%5E2%20%2B%2030x%20%2B%20200)
<h3>The length of the addition</h3>
The area of the addition is given as:
![Area = x^2 + 10x - 200](https://tex.z-dn.net/?f=Area%20%3D%20x%5E2%20%2B%2010x%20-%20200)
Expand
![Area = x^2 + 20x - 10x - 200](https://tex.z-dn.net/?f=Area%20%3D%20x%5E2%20%2B%2020x%20-%2010x%20-%20200)
Factorize
![Area = x(x + 20) - 10(x + 20)](https://tex.z-dn.net/?f=Area%20%3D%20x%28x%20%2B%2020%29%20-%2010%28x%20%2B%2020%29)
Factor out x + 20
![Area = (x -10)(x + 20)](https://tex.z-dn.net/?f=Area%20%3D%20%28x%20-10%29%28x%20%2B%2020%29)
The width of the addition is x - 10.
Hence, the length of the addition is x + 20
<h3>The area of the original house</h3>
The dimension of the original house is
x + 20 by x + 10
So, the area is:
![Area =(x + 20)(x + 10)](https://tex.z-dn.net/?f=Area%20%3D%28x%20%2B%2020%29%28x%20%2B%2010%29)
Expand
![Area =x^2 + 20x + 10x + 200](https://tex.z-dn.net/?f=Area%20%3Dx%5E2%20%2B%2020x%20%2B%2010x%20%2B%20200)
This gives
![Area =x^2 + 30x + 200](https://tex.z-dn.net/?f=Area%20%3Dx%5E2%20%2B%2030x%20%2B%20200)
Hence, the area of the original house is ![x^2 + 30x + 200](https://tex.z-dn.net/?f=x%5E2%20%2B%2030x%20%2B%20200)
Read more about areas at:
brainly.com/question/24487155