An expression is shown below:

4(m + 3 + 5m)

part a: write two expressions that are equivalent to the given expression.

part b: show that one of your expressions in part a is equivalent to the given expression using algebraic properties. explain which properties you used.

part c: show that your other expression from part a is equivalent to the given expression by substituting a number for m.

Given Expression: 4 ( m + 3 + 5m )

Part A).

To find: Two Equivalent Expression to the given expression

1st Expression = 4m + 12 + 20m

2nd Expression = 4 ( 6m + 3 )

Part B).

To get 1st expression in part A i used Distributive property multiplication over addition.

a × ( b + c ) = a × b + a × c

Consider,

4 ( m + 3 + 5m )

Now,

4 × m + 4 × 3 + 4 × 5m

= 4m + 12 + 20m

Part C).

We need to show: 4 ( m + 3 + 5m ) = 4 ( 6m + 3 )

We show equal by putting m = 1

Consider,

LHS = 4 ( m + 3 + 5m ) = 4 ( 1 + 3 + 5 × 1 ) = 4 ( 1 + 3 + 5 ) = 4 × 9 = 36

RHS = 4 ( 6m + 5 ) = 4 ( 6 × 1 + 3 ) = 4 ( 6 + 3 ) = 4 × 9 = 36

So, LHS  = RHS

Hence Proved.

b. x + 1

step-by-step explanation:

i'am smart

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'.