Solve for x: 4 over 5 x + 4 over 3 = 2x x = __________________ Write your answer as a fraction in simplest form. Use the "/" sym
bol for the fraction bar. Answer for Blank 1:
2 answers:
The answer question is x=10/9
Let's see.
4x/5 + 4/3 = 2x
First we have to make each denominator the same, so I'll multiply 4x/5 by 3/3, 4/3 by 5/5, and 2x by 15/15
Now we have 12x/15 + 20/15 = 30x/15
With everything in the same denominator we can solve the new equation of
12x + 20 = 30x
20 = 18x
10 = 9x
X= 10/9
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Answer: 20$
Step-by-step explanation: You would add 2(5)+10=y so 10+10=y 10 + 10 = 20
In this case (5,20). Hope this helps!!
so the points are, from P1 to P2, namely P1P2, and from P2 to P3, namely P2P3, and from P3 back to P1, namely P3P1.
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ P1(\stackrel{x_1}{5}~,~\stackrel{y_1}{-4})\qquad P2(\stackrel{x_2}{8}~,~\stackrel{y_2}{-3})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ P1P2=\sqrt{[8-5]^2+[-3-(-4)]^2}\implies P1P2=\sqrt{(8-5)^2+(-3+4)^2} \\\\\\ P1P2=\sqrt{3^2+1^2}\implies \boxed{P1P2=\sqrt{10}}\\\\[-0.35em] \rule{34em}{0.25pt}](https://tex.z-dn.net/?f=%20%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20P1%28%5Cstackrel%7Bx_1%7D%7B5%7D~%2C~%5Cstackrel%7By_1%7D%7B-4%7D%29%5Cqquad%20%20P2%28%5Cstackrel%7Bx_2%7D%7B8%7D~%2C~%5Cstackrel%7By_2%7D%7B-3%7D%29%5Cqquad%20%5Cqquad%20%20d%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20P1P2%3D%5Csqrt%7B%5B8-5%5D%5E2%2B%5B-3-%28-4%29%5D%5E2%7D%5Cimplies%20P1P2%3D%5Csqrt%7B%288-5%29%5E2%2B%28-3%2B4%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20P1P2%3D%5Csqrt%7B3%5E2%2B1%5E2%7D%5Cimplies%20%5Cboxed%7BP1P2%3D%5Csqrt%7B10%7D%7D%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%20)
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ P2(\stackrel{x_2}{8}~,~\stackrel{y_2}{-3})\qquad P3(\stackrel{x_2}{7}~,~\stackrel{y_2}{-10}) \\\\\\ P2P3=\sqrt{[7-8]^2+[-10-(-3)]^2}\implies P2P3=\sqrt{(7-8)^2+(-10+3)^2} \\\\\\ P2P3=\sqrt{(-1)^2+(-7)^2}\implies P2P3=\sqrt{50}\implies \boxed{P2P3=5\sqrt{2}}\\\\[-0.35em] \rule{34em}{0.25pt}](https://tex.z-dn.net/?f=%20%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20P2%28%5Cstackrel%7Bx_2%7D%7B8%7D~%2C~%5Cstackrel%7By_2%7D%7B-3%7D%29%5Cqquad%20%20P3%28%5Cstackrel%7Bx_2%7D%7B7%7D~%2C~%5Cstackrel%7By_2%7D%7B-10%7D%29%20%5C%5C%5C%5C%5C%5C%20P2P3%3D%5Csqrt%7B%5B7-8%5D%5E2%2B%5B-10-%28-3%29%5D%5E2%7D%5Cimplies%20P2P3%3D%5Csqrt%7B%287-8%29%5E2%2B%28-10%2B3%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20P2P3%3D%5Csqrt%7B%28-1%29%5E2%2B%28-7%29%5E2%7D%5Cimplies%20P2P3%3D%5Csqrt%7B50%7D%5Cimplies%20%5Cboxed%7BP2P3%3D5%5Csqrt%7B2%7D%7D%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%20)
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ P3(\stackrel{x_2}{7}~,~\stackrel{y_2}{-10})\qquad P1(\stackrel{x_1}{5}~,~\stackrel{y_1}{-4}) \\\\\\ P3P1=\sqrt{[5-7]^2+[-4-(-10)]^2}\implies P3P1=\sqrt{(5-7)^2+(-4+10)^2} \\\\\\ P3P1=\sqrt{(-2)^2+6^2}\implies P3P1=\sqrt{40}\implies \boxed{P3P1=2\sqrt{10}}](https://tex.z-dn.net/?f=%20%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20P3%28%5Cstackrel%7Bx_2%7D%7B7%7D~%2C~%5Cstackrel%7By_2%7D%7B-10%7D%29%5Cqquad%20%20P1%28%5Cstackrel%7Bx_1%7D%7B5%7D~%2C~%5Cstackrel%7By_1%7D%7B-4%7D%29%20%5C%5C%5C%5C%5C%5C%20P3P1%3D%5Csqrt%7B%5B5-7%5D%5E2%2B%5B-4-%28-10%29%5D%5E2%7D%5Cimplies%20P3P1%3D%5Csqrt%7B%285-7%29%5E2%2B%28-4%2B10%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20P3P1%3D%5Csqrt%7B%28-2%29%5E2%2B6%5E2%7D%5Cimplies%20P3P1%3D%5Csqrt%7B40%7D%5Cimplies%20%5Cboxed%7BP3P1%3D2%5Csqrt%7B10%7D%7D%20)
The valued of the expression is 1 .
4,8,12,16,20,24,28,32,36,40,44,48
Answer:
Son in law
Step-by-step explanation:
