Answer:
![\boxed{\boxed{\sf x = 32{}^{ \circ}}}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cboxed%7B%5Csf%20x%20%3D%2032%7B%7D%5E%7B%20%5Ccirc%7D%7D%7D%20)
Step-by-step explanation:
The given pair of angles is Linear pair.
<em>[</em><em>*If the adja</em><em>c</em><em>ent angles are drawn in such a way that they form </em><em>straight</em><em> </em><em>line</em> <em>,</em><em> </em><em>then </em><em>the </em><em>angles</em><em> </em><em>are </em><em>known</em><em> </em><em>as </em><em>linear</em><em> </em><em>pair*</em><em>]</em>
We know that their sum is equal to 180°.
<h3><em><u>So,</u></em></h3>
![\rm\angle{POR} + \angle{QOR} = \tt180 {}^{ \circ}](https://tex.z-dn.net/?f=%20%20%5Crm%5Cangle%7BPOR%7D%20%2B%20%20%5Cangle%7BQOR%7D%20%3D%20%20%5Ctt180%20%7B%7D%5E%7B%20%5Ccirc%7D%20)
<em>Note </em><em>that,</em>
[POR = (2x+60)°]
AND,
[QOR = (3x-40)°]
![\tt \implies(2x + 60) {}^{ \circ} + (3x - 40){}^{ \circ} = 180{}^{ \circ}](https://tex.z-dn.net/?f=%20%5Ctt%20%5Cimplies%282x%20%2B%2060%29%20%7B%7D%5E%7B%20%5Ccirc%7D%20%20%2B%20%283x%20-%2040%29%7B%7D%5E%7B%20%5Ccirc%7D%20%20%3D%20180%7B%7D%5E%7B%20%5Ccirc%7D%20)
![\tt \implies2x + 60 + 3x - 40 = 180{}^{ \circ}](https://tex.z-dn.net/?f=%5Ctt%20%5Cimplies2x%20%20%2B%2060%20%20%2B%203x%20-%2040%20%3D%20180%7B%7D%5E%7B%20%5Ccirc%7D%20)
<em>Now find the value of x.</em>
<u>Steps:</u>
![\tt \implies2x + 60 + 3x - 40 = 180{}^{ \circ}](https://tex.z-dn.net/?f=%5Ctt%20%5Cimplies2x%20%20%2B%2060%20%20%2B%203x%20-%2040%20%3D%20180%7B%7D%5E%7B%20%5Ccirc%7D%20)
<em>The LHS may be rewritten as,</em>
![\tt \implies2x + 3x + 60 + ( - 40) = 180{}^{ \circ}](https://tex.z-dn.net/?f=%5Ctt%20%5Cimplies2x%20%2B%203x%20%20%2B%2060%20%2B%20%28%20-%2040%29%20%3D%20180%7B%7D%5E%7B%20%5Ccirc%7D%20)
<u>So, Combine like terms :</u>
![\tt \implies(2x + 3x) + 60 - 40 = 180{}^{ \circ}](https://tex.z-dn.net/?f=%5Ctt%20%5Cimplies%282x%20%2B%203x%29%20%2B%2060%20-%2040%20%3D%20180%7B%7D%5E%7B%20%5Ccirc%7D%20)
![\tt \implies3x + (60 - 40) = 180 {}^ \circ](https://tex.z-dn.net/?f=%5Ctt%20%5Cimplies3x%20%2B%20%2860%20-%2040%29%20%3D%20180%20%20%7B%7D%5E%20%5Ccirc)
- <u>Subtract the bracketed portion:</u>
![\tt \implies5x + 20 = 180{}^{ \circ}](https://tex.z-dn.net/?f=%5Ctt%20%5Cimplies5x%20%20%2B%2020%20%3D%20180%7B%7D%5E%7B%20%5Ccirc%7D%20)
<u>Now, Subtract 20 from both sides:</u>
![\tt \implies5x + 20 - 20 = 180{}^{ \circ} - 20](https://tex.z-dn.net/?f=%5Ctt%20%5Cimplies5x%20%2B%2020%20-%2020%20%3D%20180%7B%7D%5E%7B%20%5Ccirc%7D%20%20-%2020)
- <u>Simplify the RHS and LHS:</u>
![\tt \implies5x + 0 = 160{}^{ \circ}](https://tex.z-dn.net/?f=%5Ctt%20%5Cimplies5x%20%20%2B%200%20%3D%20160%7B%7D%5E%7B%20%5Ccirc%7D%20)
![\tt \implies5x = 160{}^{ \circ}](https://tex.z-dn.net/?f=%5Ctt%20%5Cimplies5x%20%3D%20160%7B%7D%5E%7B%20%5Ccirc%7D%20)
<u>Divide both sides by 5:</u>
![\tt \implies \cfrac{5x}{5} = \cfrac{160{}^{ \circ} }{5}](https://tex.z-dn.net/?f=%5Ctt%20%5Cimplies%20%5Ccfrac%7B5x%7D%7B5%7D%20%20%3D%20%20%5Ccfrac%7B160%7B%7D%5E%7B%20%5Ccirc%7D%20%7D%7B5%7D%20)
- <u>Cancel the RHS and LHS:</u>
![\tt \implies \cfrac {\cancel5 {}^{1} x}{ \cancel5 {}^{1} } = \cfrac{ \cancel{160} {}^{ \circ} \: {}^{32{}^{ \circ} } }{ \cancel5 {}^{1} }](https://tex.z-dn.net/?f=%5Ctt%20%5Cimplies%20%5Ccfrac%20%7B%5Ccancel5%20%7B%7D%5E%7B1%7D%20x%7D%7B%20%5Ccancel5%20%7B%7D%5E%7B1%7D%20%7D%20%20%3D%20%20%5Ccfrac%7B%20%5Ccancel%7B160%7D%20%7B%7D%5E%7B%20%5Ccirc%7D%20%5C%3A%20%20%7B%7D%5E%7B32%7B%7D%5E%7B%20%5Ccirc%7D%20%7D%20%7D%7B%20%5Ccancel5%20%7B%7D%5E%7B1%7D%20%7D%20)
![\tt \implies1x = 32{}^{ \circ}](https://tex.z-dn.net/?f=%5Ctt%20%5Cimplies1x%20%3D%2032%7B%7D%5E%7B%20%5Ccirc%7D%20)
![\tt \implies{x} = 32{}^{ \circ}](https://tex.z-dn.net/?f=%5Ctt%20%5Cimplies%7Bx%7D%20%3D%2032%7B%7D%5E%7B%20%5Ccirc%7D%20)
Hence, the value of x would be 32°.
![\rule{225pt}{2pt}](https://tex.z-dn.net/?f=%20%5Crule%7B225pt%7D%7B2pt%7D)
I hope this helps!
Let me know if you have any questions.
The first is 2 the second one is -4
Combine like terms:
subtract x from both sides to get
2x+5=14
subtract 5
2x=9
isolate x:
divide 2 from both sides to isolate x
x=9/2
that's your final answer. unless your teacher wants you to put it in decimal form in which all you do is divide 9÷2.
Perimeter of a rectangle = 2(l + w)
Length =
![\sqrt{(1+2)^2+(0-2)^2} = \sqrt{3^2+(-2)^2} = \sqrt{9+4} = \sqrt{13}=3.61](https://tex.z-dn.net/?f=%20%5Csqrt%7B%281%2B2%29%5E2%2B%280-2%29%5E2%7D%20%3D%20%5Csqrt%7B3%5E2%2B%28-2%29%5E2%7D%20%3D%20%5Csqrt%7B9%2B4%7D%20%3D%20%5Csqrt%7B13%7D%3D3.61%20)
Width =
![\sqrt{(0-1)^2+(-1-0)^2} = \sqrt{(-1)^2+(-1)^2} = \sqrt{1+1} = \sqrt{2}=1.41](https://tex.z-dn.net/?f=%20%5Csqrt%7B%280-1%29%5E2%2B%28-1-0%29%5E2%7D%20%3D%20%5Csqrt%7B%28-1%29%5E2%2B%28-1%29%5E2%7D%20%3D%20%5Csqrt%7B1%2B1%7D%20%3D%20%5Csqrt%7B2%7D%3D1.41%20)
Perimeter = 2(3.61 + 1.41) = 2 x 5.02 = 10.04 units