simplify the following expression.

(x+1)^2-3(x+2)

a. x^2-3x+7
b. x^2-x-5
c. x^2-3x-5
d. x^2-x+3

simplify the following expression.

(5x^3y^-)

a. -30x^7y^2
b. -30x^12y^3
c. -30x^12/y^3
d. -30x^7/y^2

solve the following equation.

8+x/2 - x+3/3=4

a. x=-14
b. x=6
c. x=14
d. x=-6

solve the equation.

x^2-3x-8=x+4

a. x=-6 or x=2
b. x=1+ sqrt5 or x=1- sqrt5
c. x=1+2sqrt3 or x=1-2sqrt3
d. x=-2 or x=6

joseph is solving the equation x^2+8x-4=0 using the technique completing the square.
x^2+8x-4=0
x^2+8x+? =4+?
what number should joseph add to both sides of the equation to complete the square, and why?

a. he should add 4 to both sides because 8/2=4.
b. he should add 4 to both sides because (4/2)^2=4.
c. he should add 16 to both sides because 16/2=8.
d. he should add 16 to both sides because (8/2)^2=16.

step-by-step explanation:

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.