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Vitek1552 [10]
2 years ago
7

Determine the center for: x^2 + y^2 − 8x + 6y + 12 = 0

Mathematics
2 answers:
Anit [1.1K]2 years ago
7 0

Answer:

1/3

Step-by-step explanation:

Just divide

levacccp [35]2 years ago
3 0
your answer should be (4,-3)
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Please answer the following question. Also please answer if you actually know it.
Iteru [2.4K]

Given : f(x)= 3|x-2| -5

f(x) is translated 3 units down and 4 units to the left

If any function is translated down then we subtract the units at the end

If any function is translated left then we add the units with x inside the absolute sign

f(x)= 3|x-2| -5

f(x) is translated 3 units down

subtract 3 at the end, so f(x) becomes

f(x)= 3|x-2| -5 -3

f(x) is translated   4 units to the left

Add 4 with x inside the absolute sign, f(x) becomes

f(x)= 3|x-2 + 4| -5 -3

We simplify it and replace f(x) by g(x)

g(x) = 3|x + 2| - 8

a= 3, h = -2 , k = -8



5 0
3 years ago
Which calculation would result in the decimal point moving two spaces to the right in a decimal number?
Burka [1]

Answer:

multiplying a number by 100 would result in the decimal point moving to the right. example: 0.64 x 100 = 64. hope this helps

Step-by-step explanation:

- Zombie

6 0
3 years ago
What is the image of G for a dilation with center (0, 0) and a scale factor of 1?
Naily [24]
If the scale factor is 1, the image G' is the same as the original pont G. It is
  (1, -3)
4 0
3 years ago
Read 2 more answers
The county hospital is located at the center of a square whose sides are 3 miles wide. If an accident occurs within this square,
Dmitry_Shevchenko [17]

Answer:

3

Step-by-step explanation:

The x-coordinate of the accident is a random variable of uniform distribution with range [-1.5,1.5]. The same can be said for the y-coordinate. However, |x| and |y| are also uniform variables with range [0,1.5], since we still have the same probability on every pair of intervals with the same lenght.

Note that |X| and |Y| are independent with each other, therefore E(|x|+|y|) = E(|x|) + E(|y|) = 2*E(|x|). The last equality holds because both x and y are identically distributed, then so are |x| and |y|.

E(|x|) = \int\limits^{1.5}_0 {\frac{x}{1.5} \, dx = \frac{2}{3}* ((1.5^2)/2 - 0^2/2) = \frac{3}{2}

Therefore, E(|x| + |y|) = 2 * 3/2 = 3.

I hope that works for you!

4 0
3 years ago
This is science but pls help
Finger [1]
2 is the answer!! stream dynamite and back door :)
8 0
3 years ago
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