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MaRussiya [10]
2 years ago
6

TIMER

Mathematics
1 answer:
aleksklad [387]2 years ago
6 0

Answer:

the domain is all realz

the range is all realz

Step-by-step explanation:

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To find the x-int., let y = 0 and solve for x:  13x = 6, so x = 6/13:  (6/13, 0)

To find the y-int., let x =0 and read off the y-int:  (0, -6)
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Does the point (-4, 2) lie inside or outside or on the circle x^2 + y^2 = 25?​
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Given equation of the Circle is ,

\sf\implies x^2 + y^2 = 25

And we need to tell that whether the point (-4,2) lies inside or outside the circle. On converting the equation into Standard form and determinimg the centre of the circle as ,

\sf\implies (x-0)^2 +( y-0)^2 = 5 ^2

Here we can say that ,

• <u>Radius</u> = 5 units

• <u>Centre </u> = (0,0)

<u>Finding</u><u> </u><u>distance</u><u> between</u><u> </u><u>the </u><u>two </u><u>points</u><u> </u><u>:</u><u>-</u><u> </u>

\sf\implies Distance = \sqrt{ (0+4)^2+(2-0)^2} \\\\\sf\implies Distance = \sqrt{ 16 + 4 } \\\\\sf\implies Distance =\sqrt{20}\\\\\sf\implies\red{ Distance = 4.47 }

  • Here we can see that the distance of point from centre is less than the radius.

<u>Hence </u><u>the</u><u> </u><u>point</u><u> </u><u>lies </u><u>within</u><u> </u><u>the </u><u>circle</u><u> </u><u>.</u>

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