Answer:
I think it would be ![\frac{36}{4} -6](https://tex.z-dn.net/?f=%5Cfrac%7B36%7D%7B4%7D%20-6)
Step-by-step explanation:
stomp me if its wrong <3
![\frac{36}{4} -6 = 9-6 = 3](https://tex.z-dn.net/?f=%5Cfrac%7B36%7D%7B4%7D%20-6%20%3D%209-6%20%3D%203)
<em> hopefully this helps ~</em>
Answer:
x = 3 and y = 2
or
(3,2)
Step-by-step explanation:
first equation (LCM = 3 x 4 x 5 = 60)
(x+2)/5 - (9-x)/3 = (y - 6)/4
12(x+2) - 20(9-x) = 15(y - 6)
12x + 24 - 180 + 20x = 15y - 90
32x - 15y = -90 - 24 + 180
32x - 15y = 66
second equation:
8(x + 2) - (2x + y) = 2(x + 13)
8x + 16 - 2x - y = 2x + 26
6x - y + 16 = 2x + 26
4x - y = 10
Now you have 2 equations:
4x - y = 10 ---> y = 4x - 10
32x - 15y = 66
Substitute y = 4x - 10 into 32x - 15y = 66
32x - 15(4x - 10) = 66
32x - 60x + 150 = 66
-28x = -84
x = 3
y = 4(3) - 10
y = 12 - 10
y = 2
Solutions: x = 3 and y = 2
As an improper fraction, it is
![\frac{8}{5}](https://tex.z-dn.net/?f=%20%5Cfrac%7B8%7D%7B5%7D%20)
.
(-2,-5)
The first goal with equations like these is to get one of the variables to go away. You can do that by adding/subtracting one equation from the other. Then you solve for the one variable. Then you plug that answer into one of the original equations and solve.
If the circle has the same center as the diagonals of a square and the radius of the circle is smaller than 1/2 the diagonal of the square but larger than 1/2 the length of the side of a square, then there are 8 points of intersection -- 2 at each corner of the square.
If the radius of the circle is smaller than 1/2 the side length of the square and the center is as described above, there are no points of intersection.
If the circle is located outside the square it can have 1 tangent point or 2 intersection points depending on the location conditions of the circle in relation to the square.