What is the radius of a circle whose equation is x2+y2−10x+6y+18=0?

The radius is 4 units

Step-by-step explanation:

We can find the radius of a circle through the following procedure:

Complete the square for x and y  

Put in the format for a circle. (x-h)² + (y-k)² = r², center at (h,k) with radius = r

x2+y2−10x+6y+18=0

Now subtract 18 from both sides.

x^2+y^2−10x+6y+18-18=0-18

x^2 - 10x + y^2 + 6y = -18

(x-5)^2 - 25 + (y+3)^2 - 9 = -18

(x-5)^2 + (y+3)^2 = -18+9+25

(x-5)^2 + (y+3)^2 = 16

So,

r^2 = 16

Take square root at both sides:

√r^2 = √16

r = 4

Thus the radius is 4 units

9) i can't see the angle so whatever it is just do this: cos(angle) x hypotenuse and then use that answer and they hypotenuse and plug it into this: (answer you just got)^2 + b^2 = (hypotenuse)^2. the first answer will be the answer of the adjacent side and the second answer after this equation will be the opposite side.

10) y = 14 x = √147 or 12.12435565

11) x = 4.8

12) x = 4.2

13) x = 9.3

14) pq: 13.2

qr : 17.6

angle p: 53º

15) angle c: 54.3º

angle a: 35.7º

16) h = 21040

v = 1841

step-by-step explanation:

10) 7/sin(30) = 14 = y

7^2 + b^2 = 14^2

49 + b^2 = 196

b^2 = 196 - 49

√b^2 = √147

x = √147

11) cos(70) x 14 = 4.788282007

x = 4.8

12) tan(25) x 9 = 4.196768923

x = 4.2

13) sin(31) x 18 = 9.270685348

x = 9.3

14) pq: sin(37) x 22 = 13.23993051

pq: 13.2

qr: cos(37) x 22 = 17.56998122

qr : 17.6

angle p: 180 - (37 + 90) = 53º

15) cos-1(7/12) = 54.31466529

angle c: 54.3º

angle a: 180 - (90 + 54.31466529) = 35.68533471

angle a: 35.7º

16) 4 x 5280 = 21120

cos(5) x 21120

h = 21039.63202

h = 21040

sin(5) x 21120

v = 1840.729287

v = 1841

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step-by-step explanation: