Sixteen. For the fraction 6/10 which is equivalent to 3/5
Answer:
The solution for the given expression x ix 6.
Step-by-step explanation:
![4x + 1 = 5x + (-5)](https://tex.z-dn.net/?f=4x%20%2B%201%20%3D%205x%20%2B%20%28-5%29%20)
Solution:
Step 1: Adding (-4x) on both sides
![4x + 1 +(-4)= 5x + (-5)+(-4x)](https://tex.z-dn.net/?f=4x%20%2B%201%20%2B%28-4%29%3D%205x%20%2B%20%28-5%29%2B%28-4x%29%20)
![1=-5+1x](https://tex.z-dn.net/?f=1%3D-5%2B1x)
Step 2: Adding (5) on both sides:
![1+5=-5+1x+5](https://tex.z-dn.net/?f=1%2B5%3D-5%2B1x%2B5)
![6=x](https://tex.z-dn.net/?f=6%3Dx)
The solution for the given expression x ix 6.
5 days because there are 24 hours in a day so every six hours you can take 4 a day
By definition of absolute value, you have
![f(x) = |x+1| = \begin{cases}x+1&\text{if }x+1\ge0 \\ -(x+1)&\text{if }x+1](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%7Cx%2B1%7C%20%3D%20%5Cbegin%7Bcases%7Dx%2B1%26%5Ctext%7Bif%20%7Dx%2B1%5Cge0%20%5C%5C%20-%28x%2B1%29%26%5Ctext%7Bif%20%7Dx%2B1%3C0%5Cend%7Bcases%7D)
or more simply,
![f(x) = \begin{cases}x+1&\text{if }x\ge-1\\-x-1&\text{if }x](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cbegin%7Bcases%7Dx%2B1%26%5Ctext%7Bif%20%7Dx%5Cge-1%5C%5C-x-1%26%5Ctext%7Bif%20%7Dx%3C-1%5Cend%7Bcases%7D)
On their own, each piece is differentiable over their respective domains, except at the point where they split off.
For <em>x</em> > -1, we have
(<em>x</em> + 1)<em>'</em> = 1
while for <em>x</em> < -1,
(-<em>x</em> - 1)<em>'</em> = -1
More concisely,
![f'(x) = \begin{cases}1&\text{if }x>-1\\-1&\text{if }x](https://tex.z-dn.net/?f=f%27%28x%29%20%3D%20%5Cbegin%7Bcases%7D1%26%5Ctext%7Bif%20%7Dx%3E-1%5C%5C-1%26%5Ctext%7Bif%20%7Dx%3C-1%5Cend%7Bcases%7D)
Note the strict inequalities in the definition of <em>f '(x)</em>.
In order for <em>f(x)</em> to be differentiable at <em>x</em> = -1, the derivative <em>f '(x)</em> must be continuous at <em>x</em> = -1. But this is not the case, because the limits from either side of <em>x</em> = -1 for the derivative do not match:
![\displaystyle \lim_{x\to-1^-}f'(x) = \lim_{x\to-1}(-1) = -1](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bx%5Cto-1%5E-%7Df%27%28x%29%20%3D%20%5Clim_%7Bx%5Cto-1%7D%28-1%29%20%3D%20-1)
![\displaystyle \lim_{x\to-1^+}f'(x) = \lim_{x\to-1}1 = 1](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bx%5Cto-1%5E%2B%7Df%27%28x%29%20%3D%20%5Clim_%7Bx%5Cto-1%7D1%20%3D%201)
All this to say that <em>f(x)</em> is differentiable everywhere on its domain, <em>except</em> at the point <em>x</em> = -1.
You need to move the decimal point to be at 1.
Since this is a decimal, the power will be negative.
I had to move the decimal 6 places.
So,
1^-6
I hope this helps!
~kaikers