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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A line parallel to y = 0.6x + 3 will be of the form:

where k is a constant.
It passes through (-3,-5)
Thus,

Answer: none
Step-by-step explanation:
(A)
(16÷32/10) ×2 + 0.2×(90)
Using bodmas principle ; solve bracket
(16×10/32)×2 + (2/10×90)
10+18 =28
(B)
{(16÷32/10) × (2+2/10)} ×90
Open brackets
{(16×10/32) × (22/10)} ×90
(5×11/5) ×90
11×90 = 990
(C)
16÷{(32/10×2) + (2/10×8)} +82
Open brackets, solve division first, dolled by addition
16÷(32/5 + 8/5) +82
16÷(40/5) +82
16÷8 +82
2+82= 84
(D)
[16÷(32/10 ×2) + 0.2× (90)]
16÷ (32/5) + 2/10 ×90
Solve division
16×5/32 + 18
5/2 + 18
L.c.m of denominator (2&1) =2
(5+36) / 2 = 41/2
=20.5
4hrs, I think, because 20 divided by 5 is 4.