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torisob [31]
3 years ago
8

Find the equation of circle whose centre is at (3,4) and passing through (1,5)​

Mathematics
2 answers:
kumpel [21]3 years ago
4 0

Answer:

Step-by-step explanation:

kifflom [539]3 years ago
4 0

Answer:

(x-3)^{2} + (y-4)^{2} = r^{2} \\\\(1-3)^{2} + (5-4)^{2} = r^{2}\\r^{2}  = 5 \\so the equation is \\(x-3)^{2} + (y-4)^{2} = 5

Step-by-step explanation:

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X^2 - 2x = 15 how do I solve this quadratic method?<br>I am stuck on it.... please help<br>​
Anarel [89]

Answer:

(x - 5)(x + 3)

Step-by-step explanation:

to solve x² - 2x = 15, we need to get all the terms on one side so we can solve the quadratic equation using factoring. to do this, we subtract 15 from both sides

x² - 2x = 15

- 15         -15

x² - 2x - 15 = 0

now we can factor. we need 2 numbers that when multiplied together give us -15, and when those 2 numbers are added together we get -2

because we have a -15, we can assume that one number must be negative and the other positive, as a negative times a postive is a negative.

we can use 3 and -5 as factors and test it out. we put x in front because we have an x²

(x - 5)(x + 3) < we can FOIL to check to see if this is correct. its not mandatory to check but when you arent sure of the answer you can FOIL it out

FOIL stands for: First, Outside, Inside, and Last terms

F: (x - 5)(x + 3) = x²

O: (x - 5)(x + 3) = -3x

I: (x - 5)(x + 3) = 5x

L: (x - 5)(x + 3) = -15

x² + 3x - 5x - 15 < subtract 3x from 5x

x² - 2x - 15

this checks out, so our answer is (x - 5)(x + 3)

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