Please help with 3 math questions ( 50 points ) 1. Graph ΔABC with A(5, 1), B(-2, 6) C(-3, 4), and its image after a reflection along the x-axis. What are the coordinates of A'B'C'?

A'(-5, 1), B'(2, 6), C'(3, 4)

A'(1, -5), B'(6, 2), C'(4, 3)

A'(5, -1), B'(-2, -6), C'(-3, -4)

A'(1, 5), B'(6, -2), C'(4, -3)

A'(-5, -1), B'(2, -6), C'(3, -4)

2. Graph ΔXYZ with X(1, 2), Y(9, 7) Z(10, 1), and its image after a counterclockwise rotation of 90° about the origin. What are the coordinates of X'Y'Z'?

X'(-1, 2), Y'(-9, 7), Z'(-10, 1)

X'(1, -2), Y'(9, -7), Z'(10, -1)

X'(-1, -2), Y'(-9, -7), Z'(-10, -1)

X'(2, -1), Y'(7, -9), Z'(1, -10)

X'(-2, 1), Y'(-7, 9), Z'(-1, 10)

3. Graph parallelogram WXYZ with W(2, -3), X(-5, -1), Y(-1, 2), Z(6, 0), and its image after a translation of (x, y) → (x – 5, y + 2) What are the coordinates of W'?

(-3, -5)

(-3, -1)

(-3, 1)

(-5, 1)

(3, -5)

Step-by-step explanation:

1. Reflection over the x-axis: (x, y) → (x, -y)

ΔABC with A(5, 1), B(-2, 6) C(-3, 4), after a reflection along the x-axis becomes A'(5, -1), B'(-2, -6) C'(-3, -4)

2. Counterclockwise rotation of 90° about the origin: (x, y) → (-y, x)

ΔXYZ with X(1, 2), Y(9, 7) Z(10, 1), after a counterclockwise rotation of 90° about the origin the coordinates are X'(-2,1) Y'(-7,9) Z'(-1,10)

3.

Parallelogram WXYZ with W(2, -3), X(-5, -1), Y(-1, 2), Z(6, 0), has the image after a translation of (x, y) → (x – 5, y + 2) as

W'(2-5, -3+2), X'(-5-5, -1+2), Y'(-1-5, 2+2), Z'(6-5, 0+2)

W'(-3, -1), X'(-10, 1), Y'(-6, 4), Z'(1, 2)

The coordinates of W'(-3, -1)

See Graphs...

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!