Answer:
3
7
2
5
If f(x)=2x and g(x)=x−3, then find f(x)·g(x).
1 answer:
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Check the picture below.
so, the center of the circle is the midpoint of that diametrical segment, and half that length is the radius.
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~ -2 &,& -4~) % (c,d) &&(~ 3 &,& 8~) \end{array}~~~ % distance value d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d=\sqrt{[3-(-2)]^2+[8-(-4)]^2}\implies d=\sqrt{(3+2)^2+(8+4)^2} \\\\\\ d=\sqrt{25+144}\implies d=\sqrt{169}\implies d=13\qquad\qquad \qquad \stackrel{radius}{\frac{13}{2}}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%0A%5C%5C%5C%5C%0A%5Cbegin%7Barray%7D%7Bccccccccc%7D%0A%26%26x_1%26%26y_1%26%26x_2%26%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0A%26%26%28~%20-2%20%26%2C%26%20-4~%29%20%0A%25%20%20%28c%2Cd%29%0A%26%26%28~%203%20%26%2C%26%208~%29%0A%5Cend%7Barray%7D~~~%20%0A%25%20%20distance%20value%0Ad%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%5B3-%28-2%29%5D%5E2%2B%5B8-%28-4%29%5D%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%283%2B2%29%5E2%2B%288%2B4%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B25%2B144%7D%5Cimplies%20d%3D%5Csqrt%7B169%7D%5Cimplies%20d%3D13%5Cqquad%5Cqquad%20%5Cqquad%20%20%5Cstackrel%7Bradius%7D%7B%5Cfrac%7B13%7D%7B2%7D%7D)
so, the center of the circle is the midpoint of that diametrical segment, and half that length is the radius.
Answer: A
Step-by-step explanation:
If you look only at the ranges provided for x A is the only one that fits the graph. The domain of the first function () is everything equal to or less than 2. The equal to is represented by the closed circle at point (2,5) which represents that this value is included for that function. The other function continues on with values greater than 2, but does not include the x value 2 as it has an open circle at point (2,10).