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3 years ago

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Mathematics

1 answer:

Klio2033 [76]3 years ago

5 0

Answer:

(x - 1)^2 + (y - 0.5)^2 = 25.

Step-by-step explanation:

We need to calculate the radius and the coordinates of the center of the circle.

The center is at the midpoint of the diameter = (-1 + 3)/2 , (2+-1)/2

= (1, 0.5).

The radius = 1/2 * diameter = 1/2 * √[ (-1-3)^2 + (2 - -1)^2 ]

r = √25/2 = 2.5

r^2 = 2.5^2 = 6.25.

The general form of the equation is

(x - a)^2 + (y -b)^2 = r^2 where (a,b) is the center and r = the radius.

So the required equation is:

(x - 1)^2 + (y - 0.5)^2 = 6.25.

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How do you find a vector that is orthogonal to 5i + 12j ?

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$\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad \cfrac{a}{b}\\\\
slope=\cfrac{a}{{{ b}}}\qquad negative\implies -\cfrac{a}{{{ b}}}\qquad reciprocal\implies - \cfrac{{{ b}}}{a}\\\\
-------------------------------\\\\$

$\bf \boxed{5i+12j}\implies 
\begin{array}{rllll}
\ \textless \ 5&,&12\ \textgreater \ \\
x&&y
\end{array}\quad slope=\cfrac{y}{x}\implies \cfrac{12}{5}
\\\\\\
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\\\\\\
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if we were to place <5, 12> in standard position, so it'd be originating from 0,0, then the rise is 12 and the run is 5.

so any other vector that has a negative reciprocal slope to it, will then be perpendicular or "orthogonal" to it.

so... for example a parallel to <-12, 5> is say hmmm < -144, 60>, if you simplify that fraction, you'd end up with <-12, 5>, since all we did was multiply both coordinates by 12.

or using a unit vector for those above, then

$\bf \textit{unit vector}\qquad \cfrac{\ \textless \ a,b\ \textgreater \ }{||\ \textless \ a,b\ \textgreater \ ||}\implies \cfrac{\ \textless \ a,b\ \textgreater \ }{\sqrt{a^2+b^2}}\implies \cfrac{a}{\sqrt{a^2+b^2}},\cfrac{b}{\sqrt{a^2+b^2}}
\\\\\\
\cfrac{12,-5}{\sqrt{12^2+5^2}}\implies \cfrac{12,-5}{13}\implies \boxed{\cfrac{12}{13}\ ,\ \cfrac{-5}{13}}
\\\\\\
\cfrac{-12,5}{\sqrt{12^2+5^2}}\implies \cfrac{-12,5}{13}\implies \boxed{\cfrac{-12}{13}\ ,\ \cfrac{5}{13}}$

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3 years ago

A) 5x – 6y = 25 [x, y = 10]

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Ok so what u want to 58

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