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schepotkina [342]
3 years ago
15

Write an equation for a circle with a diameter that has endpoints at (–4, –7) and (–2, –5). Round to the nearest tenth if necess

ary. Question 9 options: (x + 3)2 + (y + 6)2 = 2 (x + 3)2 + (y + 6)2 = 8 (x – 3)2 + (y – 6)2 = 2 (x – 3)2 + (y – 6)2 = 8
Mathematics
2 answers:
Zinaida [17]3 years ago
4 0

since we know the endpoints of the circle, we know then that distance from one to another is really the diameter, and half of that is its radius.

we can also find the midpoint of those two endpoints and we'll be landing right on the center of the circle.

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{[-2-(-4)]^2+[-5-(-7)]^2}\implies d=\sqrt{(-2+4)^2+(-5+7)^2} \\\\\\ d=\sqrt{2^2+2^2}\implies d=\sqrt{2\cdot 2^2}\implies d=2\sqrt{2}~\hfill \stackrel{~\hfill radius}{\cfrac{2\sqrt{2}}{2}\implies\boxed{ \sqrt{2}}} \\\\[-0.35em] ~\dotfill

\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5})\qquad \qquad \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{-2-4}{2}~~,~~\cfrac{-5-7}{2} \right)\implies \left( \cfrac{-6}{2}~,~\cfrac{-12}{2} \right)\implies \stackrel{center}{\boxed{(-3,-6)}} \\\\[-0.35em] ~\dotfill

\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-3}{ h},\stackrel{-6}{ k})\qquad \qquad radius=\stackrel{\sqrt{2}}{ r} \\[2em] [x-(-3)]^2+[y-(-6)]^2=(\sqrt{2})^2\implies (x+3)^2+(y+6)^2=2

Sonja [21]3 years ago
4 0

Answer:

FIRST OPTION: (x+3)^2+ (y+6)^2 =2

Step-by-step explanation:

The equation of the circle in center-radius form is:

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

Where the center is at the point (h, k) and the radius is "r".

We know that the endpoints of the diameter of this circle are (-4, -7) and (-2, -5), so we can find the radius and the center of the circle.

In order to find the radius, we need to find the diameter. To do this, we need to use the formula for calculate the distance between two points:

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

Then, substituting the coordinates of the endpoints of the diameter into this formula, we get:

d=\sqrt{(-4-(-2))^2+(-7-(-5))^2}=2\sqrt{2}

Since the radius is half the diameter, this is:

r=\frac{2\sqrt{2}}{2}=\sqrt{2}

To find the center, given the endpoints of the diameter, we need to find the midpoint with this formula:

M=(\frac{x_2+x_1}{2},\frac{y_2+y_1}{2})

This is:

M=(\frac{-4-2}{2},\frac{-7-5}{2})=(-3,-6)

Then:

h=-3\\k=-6

Finally, substituting values  into  (x- h)^2 + (y- k)^2 = r^2, we get the following equation:

 (x- (-3))^2 + (y- (-6))^2 = (\sqrt{2})^2

(x+3)^2+ (y+6)^2 =2

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barxatty [35]

Answer:

1/8 ;

12 / 51

Step-by-step explanation:

To select any 5 numbers out of 40

Number of selection required / total numbers

Probability of winning = 5 /40 = 1/8

Total number of socks = 18

White, W = 8

Black, B = 6

Blue, C = 4

probability, without looking in the drawer, that you will first select and remove a black pair, then select either a blue or a white pair?

P(B) * [(P(C) + P(W)]

Probability = required outcome / Total possible outcomes

P(B) = 6 / 18 = 1/3

Without replacement :

P(W or C) = 8/17 + 4/17 = 12/17

Hence,

1/3 * 12/17

= 12 / 51

8 0
3 years ago
A cash box of $1 and $5 bills is worth $45. The number of $1 bills is three more than the number of $5 bills. How many of each b
Softa [21]

Answer:

  • 7 $5 bills
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Step-by-step explanation:

If we take out the extra $3, we can group the bills into one each of $5 and $1, for a value of $6. There will be 7 such groups in the remaining $42.

That means there are 7 bills of the $5 denomination, and 3 more than that (10 bills) of the $1 denomination.

There are 7 $5 bills and 10 $1 bills.

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If you want to write an equation, it is usually best to let a variable stand for the most-valuable contributor. Here, we can let x represent then number of $5 bills. Then the value of the cash box is ...

  5x +(x+3) = 45

  6x = 42 . . . . . . . . subtract 3, collect terms

  x = 7 . . . . . . . . . . . there are 7 $5 bills

  x+3 = 10 . . . . . . . . there are 10 $1 bills

You may notice that this working parallels the verbal description above. (After we subtract $3, x is the number of $6 groups.)

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Solving for x, we get

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7 0
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