2.5x - 3 + 2x = 1.75x - 1.25x + 13
4.5x - 3 = 0.50x + 13
4.0x = 16
x = 4
Answer: 4
2. We'll assume those xs are in the numerator
(3/7)x - 1/4 + (3/7)x = (9/7)x - (3/7)x + 3/4 - 1
(6/7)x - 1/4 = (6/7)x - 1/4
0 = 0
That's always true.
Answer: Any x is a solution.
The fifth square root as in a^(1/2)^(1/2)^(1/2)^(1/2)^(1/2)
Well that is equal to a^((1/2)^5) or a^(1/32)
Since a=x^16 in this case and the rule (b^a)^c=b^(a*c) we have:
(x^16)^(1/32)
x^(16/32)
x^(1/2) or if you prefer
√x
Answer:
Step-by-step explanation:6 type
5/25=1/5=20%
1-1/5=4/5=80%
Answer:58
Step-by-step explanation:
Check the picture below.
first off we'll need to know the midpoints of ST and UV, well, from ST as you can see in the picture, is just a vertical line, so its midpoint is simply (-2,0), let's find it for UV
![\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ U(\stackrel{x_1}{3}~,~\stackrel{y_1}{-2})\qquad V(\stackrel{x_2}{13}~,~\stackrel{y_2}{10}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{13+3}{2}~~,~~\cfrac{10-2}{2} \right)\implies \left( \cfrac{16}{2}~~,~~\cfrac{8}{2} \right)\implies (8~~,~~4) \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bmiddle%20point%20of%202%20points%20%7D%20%5C%5C%5C%5C%20U%28%5Cstackrel%7Bx_1%7D%7B3%7D~%2C~%5Cstackrel%7By_1%7D%7B-2%7D%29%5Cqquad%20V%28%5Cstackrel%7Bx_2%7D%7B13%7D~%2C~%5Cstackrel%7By_2%7D%7B10%7D%29%20%5Cqquad%20%5Cleft%28%5Ccfrac%7B%20x_2%20%2B%20x_1%7D%7B2%7D~~~%20%2C~~~%20%5Ccfrac%7B%20y_2%20%2B%20y_1%7D%7B2%7D%20%5Cright%29%20%5C%5C%5C%5C%5C%5C%20%5Cleft%28%5Ccfrac%7B13%2B3%7D%7B2%7D~~%2C~~%5Ccfrac%7B10-2%7D%7B2%7D%20%5Cright%29%5Cimplies%20%5Cleft%28%20%5Ccfrac%7B16%7D%7B2%7D~~%2C~~%5Ccfrac%7B8%7D%7B2%7D%20%5Cright%29%5Cimplies%20%288~~%2C~~4%29%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-2}~,~\stackrel{y_1}{0})\qquad (\stackrel{x_2}{8}~,~\stackrel{y_2}{4})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d = \sqrt{[8-(-2)]^2+[4-0]^2}\implies d=\sqrt{(8+2)^2+(4-0)^2} \\\\\\ d=\sqrt{10^2+4^2}\implies d=\sqrt{116}\implies d\approx 10.77](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20%28%5Cstackrel%7Bx_1%7D%7B-2%7D~%2C~%5Cstackrel%7By_1%7D%7B0%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B8%7D~%2C~%5Cstackrel%7By_2%7D%7B4%7D%29%5Cqquad%20%5Cqquad%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%20d%20%3D%20%5Csqrt%7B%5B8-%28-2%29%5D%5E2%2B%5B4-0%5D%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%288%2B2%29%5E2%2B%284-0%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20d%3D%5Csqrt%7B10%5E2%2B4%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B116%7D%5Cimplies%20d%5Capprox%2010.77)
