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GiuliAzevedo
19.08.2019 •
Mathematics
Find the center of a circle with the equation: x2+y2−32x−60y+1122=0
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Ответ:
(16, 30)
Step-by-step explanation:
First the equation of a circle is:
(x-h)^2 + (y-k)^2 = r^2, where (h,k) - the center.
We rewrite the equation and set them equal :
(x-h)^2 + (y-k)^2 - r^2 = x^2+y^2− 32x − 60y +1122=0
x^2 - 2hx + h^2 + y^2 - 2ky + k^2 - r^2 = x^2 + y^2 - 32x - 60y +1122 = 0
We solve for each coeffiecient meaning if the term on the LHS contains an x then its coefficient is the same as the one on the RHS containing the x or y.
-2hx = -32x => h = 16.
-2ky = -60y => k = 30. => the center is at (16, 30)
Ответ:
70 miles /hour = r
Step-by-step explanation:
d=rt
Substitute d=350 miles and t = 5 hours
350 miles = r* 5 hours
Divide each side by 5 hours
350 miles = 5 hours = r
70 miles /hour = r