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maria [59]
3 years ago
6

What is the center of the circle that has a diameter whose endpoints are (9, 7) and (-3, -5)?

Mathematics
2 answers:
Naddika [18.5K]3 years ago
8 0

Answer:

(3, 1)

Step-by-step explanation:

Midpoint of the diameter is the center

Since endpoints are (9, 7) and (-3, -5)

Midpoint will be:

  • ((9 - 3)/2, (7-5)/2) = (3, 1)

So the center is (3, 1)

zmey [24]3 years ago
4 0

Answer:

(3, 1)

Step-by-step explanation:

Midpoint of the diameter is the center

Since endpoints are (9, 7) and (-3, -5)

Midpoint will be:

((9 - 3)/2, (7-5)/2) = (3, 1)

or 3/1, 2/2 = (3,1)

So the center is (3, 1)

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Which equation can be used to represent "three minus the difference of a number and one equals one-half of the difference of thr
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REY [17]
The answer is: x = 7 - √53      or      x = 7 + √53

The general quadratic equation is: ax² + bx + c  = 0.
But, by completing the square we turn it into: a(x + d)² + e = 0, where:<span>
d = b/2a
e = c - b²/4a
Our quadratic equation is x² - 14x -4 = 0, which is after rearrangement:
So, a = 1, b = -14, c = -4
Let's first calculate d and e:
d = b/2a = -14/2*1 = -14/2 = -7
e = c - b²/4a = -4 - (-14)</span>²/4*1 = -4 - 196/4 = -4 - 49 = -53<span>
By completing the square we have:
a(x + d)² + e = 0

1(x + (-7))</span>² + (-53) = 0
(x - 7)² - 53 = 0
(x - 7)² = 53
x - 7 = +/-√53
x = 7 +/- √53

Therefore, the solutions are:
x = 7 - √53
or
x = 7 + √53
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