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arlik [135]
3 years ago
5

Find the equation in general form of the circle with center at the origin and radius equal 6.

Mathematics
1 answer:
____ [38]3 years ago
3 0
\bf \textit{equation of a circle}\\\\ 
(x-{{ h}})^2+(y-{{ k}})^2={{ r}}^2
\qquad 
\begin{array}{lllll}
center\ (&{{ h}},&{{ k}})\qquad 
radius=&{{ r}}\\
&0&0&6
\end{array}\\\\
-------------------------------\\\\
(x-0)^2+(y-0)^2=6^2\implies x^2+y^2=36
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Tobeg says the equation 3x + 2y= 270 and y=4/5x + 20 intersect at the point (50,60) is this true how do you know
il63 [147K]

ANSWER

Yes it is very true


<u>EXPLANATION</u>

If the two equations intersect at (50,60) then this point must satisfy the two equations.


3x+2y=270---(1)

We substitute  (50,60)  in to erquation (1)


3(50)+2(60)=270


150+120=270

270=270




y=\frac{4}{5}x+20 ---(2)


We now substitute  (50,60)  in to erquation (2) also


60=\frac{4}{5}(50)+20


60=40+20


60=60


Since the point satisfy all the two equations, it is true that they intersect at (60,50)




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