


Alternatively, recall that if

, then

, and so

Take

, so that

, and we have the original limit. So the limit is equivalent to the value of

, i.e.
Answer:
For your first answer it is because they are two different ways to solve the equation
Step-by-step explanation:
For Example!
<h2>Solution 1:</h2>
(x-4)² – 28 = 8
The 8 is positive making it a different equation (this is like absolute value). (I am assuming you know the answer to the probelm)
<h2>Solution 2:</h2>
(x-4)² – 28 = -8
The 8 could be negative meaning that when you add the 28 to the right side it is -8+28 which will be a negative!
<h3>I hope this helps</h3><h2>IF IT DOES HELP PLEASE GIVE IT A BRAINLIEST!</h2>
PhotosRemaining(x) = 745 pictures - (20 pictures/day)*(x), for x≥0
You find the measure of angle BEC, by adding angle B and angle C which is 43+90, and then subtract that from 180 and get 180-(43+90). Then you take the measure of angle BEC (which is the same as AEB), take AEB and A which is 90 degrees and add those together. Then you take that and subtract it from 180, which is 180-(AEB+90)