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Alexxandr [17]
3 years ago
9

Please help me, circle chapter. ​

Mathematics
1 answer:
Ad libitum [116K]3 years ago
5 0

Answer:  p = 4, r = 4

<u>Step-by-step explanation:</u>

The equation of a circle is:   (x - h)² + (y - k)² = r²     where

  • (h, k) is the center of the circle
  • r is the radius of the circle

Complete the square:

x² - 10x +  _____   +     y² + py + _____     = -13 + ___ + ___

x² - 10x  + <u> (-10/2)² </u> +     y² + py + <u> (p/2)²</u>     = -13 + <u> (-10/2)² </u> + <u> (p/2)²</u>        

                  (x - 5)²  +     (y + p/2)²                = -13 + 25 + (p/2)²

                        ↓                   ↓                          |_____<u>↓</u>_____|      

                      h = 5              k = -p/2                            r²

Given: (h, k) = (5, -2)

                                          -2 = -p/2

                                          -4 = -p

                                            4 = p

   r² =  -13 + 25 + (p/2)²

       = -13 + 25 + (4/2)²

       = -13 + 25 + 4

       = 16

\sqrt{r^2}=\sqrt{16}

  r  =  4

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Using the given points, determine the slope. (0, 32) and (100, 212)
Lerok [7]

Slope = (y2 - y1)/(x2 - x1)

Slope = (212 - 32)/(100 - 0)

Slope = 180 / 100

Slope = 18/10

Slope = 9/5

Answer

D. slope = 9/5



4 0
3 years ago
Sin∅=√3-1/2 find approximate value of sec∅(sec∅+tan∅)/1+tan²∅​
Neko [114]

Answer:

The approximate value of f(\theta) = \frac{\sec \theta \cdot (\sec \theta+\tan \theta)}{1+\tan^{2}\theta} is 1.366.

Step-by-step explanation:

Let f(\theta) = \frac{\sec \theta \cdot (\sec \theta+\tan \theta)}{1+\tan^{2}\theta}, we proceed to simplify the formula until a form based exclusively in sines and cosines is found. From Trigonometry, we shall use the following identities:

\sec \theta = \frac{1}{\cos \theta} (1)

\tan\theta = \frac{\sin\theta}{\cos \theta} (2)

\cos^{2}+\sin^{2} = 1 (3)

Then, we simplify the given formula:

f(\theta) = \frac{\left(\frac{1}{\cos \theta} \right)\cdot \left(\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta}\right) }{1+\frac{\sin^{2}\theta}{\cos^{2}\theta} }

f(\theta) = \frac{\left(\frac{1}{\cos^{2} \theta} \right)\cdot (1+\sin \theta)}{\frac{\sin^{2}\theta + \cos^2{\theta}}{\cos^{2}\theta} }

f(\theta) = \frac{\left(\frac{1}{\cos^{2}\theta}\right)\cdot (1+\sin \theta)}{\frac{1}{\cos^{2}\theta} }

f(\theta) = 1+\sin \theta

If we know that \sin \theta =\frac{\sqrt{3}-1}{2}, then the approximate value of the given function is:

f(\theta) = 1 +\frac{\sqrt{3}-1}{2}

f(\theta) = \frac{\sqrt{3}+1}{2}

f(\theta) \approx 1.366

5 0
3 years ago
If M is the midpoint of AB, find the coordinate of point B if the coordinate of point A are (7,10) and the midpoint, M is(6,1)
xeze [42]

Answer:

(5, - 18)

Step-by-step explanation:

IN MIDPOINT, WE SUM BOTH X VALUES AND Y VALUES AND DIVIDE BY TWO.

(x+7)/2=6. (y+20)/2=1

X+7=12. Y+20=2

X=12-7. Y=2-20

X=5. Y=-18

B=(5, - 18)

8 0
4 years ago
Help please!<br> Select the correct answer.
zhenek [66]
The answer to your problem is letter B
5 0
2 years ago
Read 2 more answers
Awnser okease quicxkly
Strike441 [17]

Answer:

P': (-9,10)

Q': (-2,10)

R':(-2,0)

S':(-9,0)

4 0
2 years ago
Read 2 more answers
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