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dezoksy [38]
3 years ago
15

HELP ME PLEASEEEE! Thank you

Mathematics
1 answer:
Sergio039 [100]3 years ago
6 0

Answer:

A. Yes

B. Yes

C. No

Step-by-step explanation:

We can substitute each value in to the equation and see if the sides match up. Let's start with a.

4\cdot(2)-3 = -2\cdot(2)+9\\8-3 = -4+9\\5 = 5

So, n = 2 works for equation a. Let's try B.

9\cdot(\frac{10}{3}) - 19 = 3\cdot(\frac{10}{3}) + 1\\\\\\\frac{90}{3} -19 = \frac{30}{3} + 1 \\  30 - 19 = 10 + 1\\11 = 11

So, m = \frac{10}{3} works for B. Now let's try C.

3(30+8) = 2\cdot(30)-6\\3(38) = 60-6\\114 \neq 54

So y = 30 doesn't work for C.

Hope this helped!

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25 POINTS AVAILABLE
myrzilka [38]

Answer:

\large\boxed{1.\ (-3, 0),\ r = 3}\\\boxed{2.\ (x+4)^2+(y-3)^2=36}\\\boxed{3.\ (x-2)^2+(y-1)^2=(\sqrt{34})^2}

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

1. We have the equation:

(x+3)^2+y^2=9\\\\(x-(-3))^2+(y-0)^2=3^2

<h2 />

2. We have the center (-4, 3) and the radius r = 6. Substitute:

(x-(-4))^2+(y-3)^2=6^2\\\\(x+4)^2+(y-3)^2=36

3. 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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