Answer: A = 58
Step-by-step explanation: The sketched region enclosed by the curves and the approximating rectangle are shown in the attachment.
From the sketches, the area will be integrated with respect to y.
To calculate the integral, first determine the limits, which will be the points where both curves meet.
In respect to y:
Finding limits:
Multiply by 2 to facilitate calculations:
Resolving quadratic equation:
y = 6 and y = -8
Then, integral to calculate area will be with limits -8<y<6:
![A = 24.6 - \frac{6^{3}}{6}-\frac{6^{2}}{2}-[24.(-8) - \frac{(-8)^{3}}{6}-\frac{(-8)^{2}}{2}]](https://tex.z-dn.net/?f=A%20%3D%2024.6%20-%20%5Cfrac%7B6%5E%7B3%7D%7D%7B6%7D-%5Cfrac%7B6%5E%7B2%7D%7D%7B2%7D-%5B24.%28-8%29%20-%20%5Cfrac%7B%28-8%29%5E%7B3%7D%7D%7B6%7D-%5Cfrac%7B%28-8%29%5E%7B2%7D%7D%7B2%7D%5D)
A = 58
<u>The area of the enclosed region is 58 square units.</u>
Answer:
(x, y) = (5, 7)
Step-by-step explanation:
It can be useful to put these equations into standard form, with mutually-prime coefficients and a positive x-coefficient.
10x +y -(4x +4y) = 9 . . . . . subtract the variable terms on the right
6x -3y = 9 . . . simplify
2x -y = 3 . . . . . divide by 3 to put in standard form
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(x +10y) -(10x +y) = 18 . . . . . subtract the variable terms on the right
-9x +9y = 18 . . . simplify
x -y = -2 . . . . . . . divide by -9 to put in standard form
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Now, we can subtract the second equation from the first.
(2x -y) -(x -y) = 3 -(-2)
x = 5
y = x+2 = 7 . . . from the second simplified equation
The solution is (x, y) = (5, 7).
Answer:
x^12
Step-by-step explanation:
We know that a^b* a^c = a^(b+c)
x^5 * x^7 = x^(5+7) = x^12
Answer:
(3/2,10)
Step-by-step explanation:
Mid point is ((10-7)/2,(13+7)/2)=(1.5,10)
For part a,
y < x
y > -2x
Point B: (3,1) plugin x=3, y=1 y < x 1 < 3
which is true, so point B satisfies the first inequalityYou can graph the inequality y > 2x+5 and see which schools are in the shaded region
