HELP ME!!!
1 answer:
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.
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Answer:
Step-by-step explanation:
2x + 17.6 = 6
Subtracting both sides by 17.6
2x = 6 - 17.6
2x = -11.6
Dividing both sides by 2
x = -11.6 / 2
x = -5.8
<em>Note: Since you missed to add the figure. So, after a little research I am kind of able to find the figure and hence, assuming it as a reference. Hence, I am attaching the figure and solution of that figure is also visible in the same figure.</em> Please check the attached figure (a)
Answer:
Segment B'D' will be parallel to segment BD and will be shorter than segment BD. Please see the attached figure (a) for better understanding.
Step-by-step explanation:
As we know that there are certain rules when a quadrilateral is dilated by a certain scale centered around a certain point.
So,
Let suppose P(a, b) is the point.
The dilation rule by a scale factor of 1 over 3 or (1/3) centered around (1, 2) is
P(a, b) →
Hence,
A(1, 2) → A'(1, 2)
B(2, 3) → B'(4/3, 7/3)
C(4, 2) → C'(2, 2)
D(2, 1) → D'(4/3, 5/3)
The attached figure show that if we draw the all these points in the coordinate plan i.e. A(1, 2), B(2, 3), C(4, 2), D(2, 1) and A'(1, 2), B'(4/3, 7/3), C'(2, 2), D'(4/3, 5/3), we can determine that Segment B'D' will be parallel to segment BD and will be shorter than segment BD.
<em>Keywords: dilation, quadrilateral, segment</em>
<em> Learn more about dilation from brainly.com/question/7569824</em>
<em> #learnwithBrainly</em>
