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Sphinxa [80]
3 years ago
9

Adding and subtracting function if (x)=4x^2+1and g(x)=x^2-5, find (f+g) (x)

Mathematics
1 answer:
konstantin123 [22]3 years ago
5 0

Answer:

  (f+g)(x) = 5x^2 -4

Step-by-step explanation:

  (f+g)(x)=f(x)+g(x)\\\\=(4x^2+1)+(x^2-5)=(4+1)x^2+(1-5)\\\\\boxed{(f+g)(x)=5x^2-4}

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\bf \textit{equation of a circle}\\\\ &#10;(x- h)^2+(y- k)^2= r^2&#10;\qquad &#10;center~~(\stackrel{}{ h},\stackrel{}{ k})\qquad \qquad &#10;radius=\stackrel{}{ r}\\\\&#10;-------------------------------\\\\&#10;(x+1)^2+y^2=36\implies [x-(\stackrel{h}{-1})]^2+[y-\stackrel{k}{0}]^2=\stackrel{r}{6^2}~~~~&#10;\begin{cases}&#10;\stackrel{center}{(-1,0)}\\&#10;\stackrel{radius}{6}&#10;\end{cases}

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\bf ~~~~~~~~~~~~\textit{distance between 2 points}&#10;\\\\&#10;(\stackrel{x_1}{-1}~,~\stackrel{y_1}{0})\qquad &#10;A(\stackrel{x_2}{-1}~,~\stackrel{y_2}{1})\qquad \qquad &#10;%  distance value&#10;d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}&#10;\\\\\\&#10;\stackrel{distance}{d}=\sqrt{[-1-(-1)]^2+(1-0)^2}\implies d=\sqrt{(-1+1)^2+1^2}&#10;\\\\\\&#10;d=\sqrt{0+1}\implies d=1

well, the distance from the center to A is 1, namely is "inside the circle".

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notice, C is farther than the radius 6, meaning is outside the circle, hiking about on the plane.
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