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January 30, 2015

January 30, 2015

Posted by **emer** on Saturday, September 9, 2006 at 4:14pm.

1.find the points of interesction of the line L and the circle k in the following

L:x-2y=0

K:X Squared + y squared=25

I would solve both equations for y, and then you'll have a system with two equations and two variables:

y = x/2

y = (plusorminus) sqrt(25 - x^2)

Solve the system and you'll have your points of intersection.

Use the substitution method to solve these equations. This is not a linear system because of the exponents in K.

We hve

L: x-2y=0

K: x^2+y^2=25

L is a line with slope 1/2 that goes through the origin.

K is a circle of radius 5 centered at the origin.

From L we have x=2y. Using this for x in K we have

(2y)^2+y^2=25 so 4y^2+y^2=5y^2=25 or,

y=+/-sqrt(5)

Use those values of y in either L or K to solve for the corresponding x values.

Be sure to check the answers by substituting in L and K.

I need help with this problem:

3y=15x-12

Please give me the answer ASAP. ;)

I need help with this problem:

3y=15x-12

Please give me the answer ASAP. ;)

I need help with this problem:

3y=15x-12

Please give me the answer ASAP. ;)

I need help with this problem:

3y=15x-12

Please give me the answer ASAP. ;)

i don't understand the equation?

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