The solution to this equation is in the picture i’ve put below!
By definition of absolute value, you have

or more simply,

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,

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:


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.
Answer:
12 measures
Step-by-step explanation:
Step 1:
64 : 8 Ratio
Step 2:
64 : 8 = 96 : x Equation
Step 3:
64x = 768 Multiply
Step 4:
x = 768 ÷ 64 Divide
Answer:
x = 12
Hope This Helps :)
<span>(2x – 1)(x + 7) = 0
2x-1=0
2x=1
x=1/2
x=0.5
x+7=0
x= -7
</span>
Step-by-step explanation:
first fish tank: 95-4x
second fish tank: 40+5x
1. Define variable
The variable x represents the number of days.
2. Write the inequality
<em>95-4x=40+5x</em>
<em />
3. You probably don't need the answer but I am just going to solve.
The first fish tank will have less water than the second fish tank after 7 days.
<em>95-4x=40+5x </em>
<em>95-4(7)=40+5(7)</em>
<em>95-28=40+35</em>
<em>67=75</em>
<em>Since the first equation (95-4x) represents the first fish tank, and the second equation (40+5x) represents the second fish tank, the solution shows how the amount of water in the first fish tank is less than the amount of water un the second fish tank after 7 days.</em>