Answer:
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Step-by-step explanation:
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.
Hi there! :)
A radical equation is an equation in which a variable (ex. x, y, a, v) is under a radical sign. The only equation that has a variable under a radical (x) is the second option. There is an x next to the 2 under the radical.
I hope this helps, DM me if you have anymore questions. :)
~kaikers
The four quadrants and the sign of coordinates is given as follows:
Quadrant I : (n, n ) both x and y positive.
Quadrant II : ( -n,n) = x negative and y positive.
Quadrant III : ( -n,-n) = both x and y negative.
Quadrant IV: (n, -n) = x positive and y negative.
So Based on the above concept , we can say that vertex C (-n,-n) has both x and y negative and so it lies in quadrant III.
Answer is Vertex C (-n,-n)
Answer:
A.
Step-by-step explanation:
If you use the vertical line test only one point hits the graph.