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3 years ago

I need help with #11

Mathematics

1 answer:

strojnjashka [21]3 years ago

4 0

You can plug in the numbers
1) (0,3):
3 < 2(0) + 2
3 < 2 this is incorrect, so it is not 1

2) (2,0)
0 < 2(2) + 2
0 < 6 this passed the first one

0 > -(2) - 1
0 > -3 this is also correct

the answer is 2) (2,0)

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Answer:

$\large\boxed{(x-2)^2+(y-1)^2=34}$

Step-by-step explanation:

The equation of a circle in standard form:

$(x-h)^2+(y-k)^2=r^2$

(h, k) - center

r - radius

We have the endpoints of the diameter: (-1, 6) and (5, -4).

Midpoint of diameter is a center of a circle.

The formula of a midpoint:

$\left(\dfrac{x_1+x_2}{2};\ \dfrac{y_1+y_2}{2}\right)$

Substitute:

$h=\dfrac{-1+5}{2}=\dfrac{4}{2}=2\\\\k=\dfrac{6+(-4)}{2}=\dfrac{2}{2}=1$

The center is in (2, 1).

The radius length is equal to the distance between the center of the circle and the endpoint of the diameter.

The formula of a distance between two points:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Substitute the coordinates of the points (2, 1) and (5, -4):

$r=\sqrt{(5-2)^2+(-4-1)^2}=\sqrt{3^2+(-5)^2}=\sqrt{9+25}=\sqrt{34}$

Finally we have:

$(x-2)^2+(y-1)^2=(\sqrt{34})^2$

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QUESTION :-

what is the value of x to the nearest whole number?

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