What is the remainder when x^3 -1 is divided by (x+2)

Mathematics

1 answer:

Gnom [1K]2 years ago

6 0

Answer:

Use m athway

Step-by-step explanation:

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Write an equation for the circle with endpoints of a diameter (9, 4) and (-3, -2)

inysia [295]

If the diameter is at (9,4) and (-3,-2), then the center of the circle is the midpoint of that segment.

and since we know that the radius of a circle is half of the diameter, whatever long that diameter segment is, the radius is half that.

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points }
\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~ 9 &,& 4~) 
% (c,d)
&&(~ -3 &,& -2~)
\end{array}\qquad
% coordinates of midpoint 
\left(\cfrac{ x_2 + x_1}{2}\quad ,\quad \cfrac{ y_2 + y_1}{2} \right)
\\\\\\
\left( \cfrac{-3+9}{2}~~,~~\cfrac{-2+4}{2} \right)\implies \stackrel{center}{(3~,~1)}$

$\bf -------------------------------\\\\
~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~ 9 &,& 4~) 
% (c,d)
&&(~ -3 &,& -2~)
\end{array}
\\\\\\
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
d=\sqrt{(-3-9)^2+(-2-4)^2}\implies d=\sqrt{(-12)^2+(-6)^2}
\\\\\\
d=\sqrt{180}\qquad \qquad \qquad radius=\cfrac{\sqrt{180}}{2}$

$\bf -------------------------------\\\\
\textit{equation of a circle}\\\\ 
(x- h)^2+(y- k)^2= r^2
\qquad 
center~~(\stackrel{3}{ h},\stackrel{1}{ k})\qquad \qquad 
radius=\stackrel{\frac{\sqrt{180}}{2}}{ r}
\\\\\\
(x-3)^2+(y-1)^2=\left( \frac{\sqrt{180}}{2} \right)^2\implies (x-3)^2+(y-1)^2=\cfrac{(\sqrt{180})^2}{2^2}
\\\\\\
(x-3)^2+(y-1)^2=\cfrac{180}{4}\implies (x-3)^2+(y-1)^2=45$

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