Help math, 9th grade? Please
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Help math, 9th grade? Please

[From: ] [author: ] [Date: 12-02-27] [Hit: ]
and down 4. The resulting vector can be drawn as the hypotenuse of a right triangle, if you draw in the x- and y-components as well. 3^2 + 4^2 = 5^2; the answer would be 5 (magnitude cant be negative).......
Okay so i am having some trouble with a few problems.

What is the radius of the circle whose equation is x^2 + y^2 +18x + 12y + 36=0
1)9
2)3*radical 17
3)81
4)153

Next Problem:

What is the magnitude of the vector resulting from graphing the complex number 3-4i?

1) -5
2) radical -5
3) radical 5
4) 5

Please help even if you can only answer 1 of them, i would greatly appreciate it!

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1. Complete the square for x and y. You should be able to do that, right?:

x^2 + 18x + y^2 + 12y = -36

x^2 + 18x + 81 + y^2 + 12y + 36 = 81

(x + 9)^2 + (y + 6)^2 = 81

The radius is 9, so it is 1.

2. To find the magnitude of a vector, use the Pythagorean Theorem:

sqrt (3^2 + (-4^2))

sqrt (9 + 16)

sqrt 25 = 5

The magnitude is 5.

Miss Kristin

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The equation of a circle can be written as (x + the x-coordinate of the center)^2 + (y + the y-coordinate of the center)^2 = (radius)^2. That equation can be rewritten as x^ + 18x + 81 + y^2 + 12y + 36 = 81 (you have to add 81 to both sides to complete the square for the x-coordinate), which can be simplified to (x + 9)^2 + (y + 6)^2 = 81; the radius is the square root of 81, or 9.

To graph 3 - 4i, you'd go to the right 3 on the complex number plane, and down 4. The resulting vector can be drawn as the hypotenuse of a right triangle, if you draw in the x- and y-components as well. 3^2 + 4^2 = 5^2; the answer would be 5 (magnitude can't be negative).
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