1. n/2-1/3-1=1/6
n/2=1/6+1/3+1
n/2=1/6+2/6+6/6
n/2=9/6=3/2
n=3/2*2=3
check: abs(3/2-1/3)-1=1/6, abs(9/6-2/6)-1= 1/6, 7/6-1=1/6, 1/6 =1/6
3 is a root
2. -(n/2-1/3)-1=1/6
-n/2+1/3-1=1/6
-n/2-2/3=1/6
-n/2=1/6+2/3
-n/2=1/6+4/6
-n/2=5/6
n=-2*5/6=-5/3=-1 2/3
check : abs((-5/3)/2-1/3) -1=1/6
abs(-5/6-1/3) -1 =1/6, abs(-5/6-2/6) -1=1/6, abs (-7/6)-1=1/6, 7/6-1=1/6, 1/6=1/6
so -5/3 or -1 2/3 is also root
so two roots 3 and -5/3(that can be written as -1 2/3)
Step-by-step explanation:
2(22/7)12²+2(22/7)(12)(18)
=2021.9
Just subtract 100 - 50 and you get x = 50
Answer: ![\sqrt[5]{y}](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7By%7D)
I realize its probably not the largest readable font. If you are having trouble reading it, it is the square root of y; however, there is a tiny little 5 in the upper left corner to indicate a fifth root. So you would read it out as "the fifth root of y"
The rule I'm using is
![x^{1/n} = \sqrt[n]{x}](https://tex.z-dn.net/?f=x%5E%7B1%2Fn%7D%20%3D%20%5Csqrt%5Bn%5D%7Bx%7D)
and the more general rule we could use is
![x^{m/n} = \sqrt[n]{x^m}](https://tex.z-dn.net/?f=x%5E%7Bm%2Fn%7D%20%3D%20%5Csqrt%5Bn%5D%7Bx%5Em%7D)
where m = 1. This rule helps convert from rational exponent form (aka fractional exponents) to radical form.
Answer: just put it in rice
Step-by-step explanation: