Answer:
Please check the explanation.
Step-by-step explanation:
Let us consider

To find the area under the curve
between
and
, all we need is to integrate
between the limits of
and
.
For example, the area between the curve y = x² - 4 and the x-axis on an interval [2, -2] can be calculated as:

=


solving


![=\left[\frac{x^{2+1}}{2+1}\right]^2_{-2}](https://tex.z-dn.net/?f=%3D%5Cleft%5B%5Cfrac%7Bx%5E%7B2%2B1%7D%7D%7B2%2B1%7D%5Cright%5D%5E2_%7B-2%7D)
![=\left[\frac{x^3}{3}\right]^2_{-2}](https://tex.z-dn.net/?f=%3D%5Cleft%5B%5Cfrac%7Bx%5E3%7D%7B3%7D%5Cright%5D%5E2_%7B-2%7D)
computing the boundaries

Thus,

similarly solving


![=\left[4x\right]^2_{-2}](https://tex.z-dn.net/?f=%3D%5Cleft%5B4x%5Cright%5D%5E2_%7B-2%7D)
computing the boundaries

Thus,

Therefore, the expression becomes



square units
Thus, the area under a curve is -10.67 square units
The area is negative because it is below the x-axis. Please check the attached figure.
Answer:
Step-by-step explanation:
x + 6y = 9
-x + 2y = -15
-------------------add
8y = - 6 <=== u see that the x terms cancel each other out
A = L x W
A = 8m^6n^3p x <span>m^3n^7
A = 8m^9n^10p
answer
C. </span>8m^9n^10p
Answer:
The answer is 72
First you find E=180-130=50
Then you add all the angles and subtract from 360
102+100+50=252
360-252=108
then you do 180-108=72
70 ft per sec going upward 3 sec from 6 ft high should reach 216 ft, but unfortunately gravity is pulling it and at some point it starts to go backward (at 2.176 second), so to compensate the upward slowing down and distance dropped, use -32.176 times 3 square times 1/2 = -144.792 which is the natural drop distance if the football is released mid air without a initial speed after 3 seconds. 216 - 144.792 = 71.208 ft high.