The area of the region bounded above by y= eˣ bounded by y = x, and bounded on the sides; x =0; and x = 1 is given as e¹ - 1.5.
<h3>What is the significance of "Area under the curve"?</h3>
This is the condition in which one process increases a quantity at a certain rate and another process decreases the same quantity at the same rate, and the "area" (actually the integral of the difference between those two rates integrated over a given period of time) is the accumulated effect of those two processes.
<h3>What is the justification for the above answer?</h3>
Area = 
= 
= e¹-(1/2-0); or
Area = e -1.5 Squared Unit
The related Graph is attached accordingly.
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Answer:
Step-by-step explanation:
7x + 8y + 5z
I put the equations in parentheses, assuming that I would be adding both of the equations together. Then, I removed them and combined the like terms. I used two calculators to check my answer, and this should be correct.
Step-by-step explanation:


Here, the Taylor approximation for a square root was applied, and O(x) stands for all negligible terms of Taylor's sum with respect to variable x.
So, 
b. For an increase of 2%, that is:


Answer:
57 cookies
Step-by-step explanation:
32/0.56= 57.14
Oh okie so cute ☺️ cute lol