Answer is 0= -2
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Draw a polygon with the given conditions in a coordinate plane of rectangle with a perimeter of 20 units
1 answer:
Question says to draw a polygon with the given conditions in a coordinate plane. It is talking about Rectangle with perimeter of 20 units.
So we can use formula of perimeter of rectangel to find it's possible dimensions.
We knwo that perimeter of the rectangle is given by 2(L+W)
where L is lenght and W is width.
Given perimeter is 20 so we get:
2(L+W)=20
L+W=10
L=10-W
So basically we can use any number for W which can be from 0 to 10 as side length can't be negative.
Let W=4
Then L=10-W=10-4=6
Hence we just have to draw a rectangle having length 6 and width 4.
So the final answer will be picture in the attached graph.
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I wrote it out on paper and uploaded the answer for you. Let me know if you have any questions.
Answer:
-2, 8/3
Step-by-step explanation:
You can consider the area to be that of a trapezoid with parallel bases f(a) and f(4), and width (4-a). The area of that trapezoid is ...
A = (1/2)(f(a) +f(4))(4 -a)
= (1/2)((3a -1) +(3·4 -1))(4 -a)
= (1/2)(3a +10)(4 -a)
We want this area to be 12, so we can substitute that value for A and solve for "a".
12 = (1/2)(3a +10)(4 -a)
24 = (3a +10)(4 -a) = -3a² +2a +40
3a² -2a -16 = 0 . . . . . . subtract the right side
(3a -8)(a +2) = 0 . . . . . factor
Values of "a" that make these factors zero are ...
a = 8/3, a = -2
The values of "a" that make the area under the curve equal to 12 are -2 and 8/3.
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<em>Alternate solution</em>
The attachment shows a solution using the numerical integration function of a graphing calculator. The area under the curve of function f(x) on the interval [a, 4] is the integral of f(x) on that interval. Perhaps confusingly, we have called that area f(a). As we have seen above, the area is a quadratic function of "a". I find it convenient to use a calculator's functions to solve problems like this where possible.
