A and D are the solutions that make the inequality true.
6^2= 36
9^2= 81
10^2= 100
-4^2= 16
-9^2+ 81
-12^2= 144
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.
Learn more about area bounded by curve:
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Answer:
B
Step-by-step explanation:
Siri said thats right
Perhaps the easiest way to find the midpoint between two given points is to average their coordinates: add them up and divide by 2.
A) The midpoint C' of AB is
.. (A +B)/2 = ((0, 0) +(m, n))/2 = ((0 +m)/2, (0 +n)/2) = (m/2, n/2) = C'
The midpoint B' is
.. (A +C)/2 = ((0, 0) +(p, 0))/2 = (p/2, 0) = B'
The midpoint A' is
.. (B +C)/2 = ((m, n) +(p, 0))/2 = ((m+p)/2, n/2) = A'
B) The slope of the line between (x1, y1) and (x2, y2) is given by
.. slope = (y2 -y1)/(x2 -x1)
Using the values for A and A', we have
.. slope = (n/2 -0)/((m+p)/2 -0) = n/(m+p)
C) We know the line goes through A = (0, 0), so we can write the point-slope form of the equation for AA' as
.. y -0 = (n/(m+p))*(x -0)
.. y = n*x/(m+p)
D) To show the point lies on the line, we can substitute its coordinates for x and y and see if we get something that looks true.
.. (x, y) = ((m+p)/3, n/3)
Putting these into our equation, we have
.. n/3 = n*((m+p)/3)/(m+p)
The expression on the right has factors of (m+p) that cancel*, so we end up with
.. n/3 = n/3 . . . . . . . true for any n
_____
* The only constraint is that (m+p) ≠ 0. Since m and p are both in the first quadrant, their sum must be non-zero and this constraint is satisfied.
The purpose of the exercise is to show that all three medians of a triangle intersect in a single point.
<u>Answer:</u>
The answer is A choice.
<u>Explanation:</u>
By looking at the graph, we notice that at the left side, the domain/x-value stops at -5 and 5.
Thus, the domain is at <u>the left end side to the right end side.</u>
The domain stops at -5 on the left side and 5 on the right side. Thus the answer is
-5 < x < 5.
The A choice.
