What series of transformations would move C to C'at (1,-4), S to S'at (2,-6) and R to R'at (4,-4)? A. Reflect over the y-axis, then reflect over the x-axis, then translate 3 units down.
B. Reflect over the x-axis, then translate 3 units down, then reflect over the y-axis.
C. Translate 5 units to the right, then reflect over the x-axis, then translate 3 units down.
D.
Translate 5 units down, then translate 5 units to the right, then reflect over the
x-axis.

answer:

given the linear inequality :

add 2y to both sides we get;

simplify:

add 12 to both sides of an equation;

simplify:

divide both sides by 2 we get;

or

the related equation for the inequality equation [1] is;

y-intercepts defined as the graph crosses the y-axis.

substitute x =0 in [2] to solve for y;

y = 6

x-intercepts defined as the graph crosses the x-axis.

substitute value y = 0 to solve for x;

subtract 6 from both sides we get

multiply both sides by 2 we get;

or

x = -12

now, plot these point (0, 6) and (-12, 0) on graph as shown below and shade the graph below the solid line.

hey there! : )

4y < -4

we need to isolate y. we can do this by simply dividing both sides by 4.

4y ÷ 4 > -4 ÷ 4

simplify.

y > -1

5 - y > 0

once again, we need to isolate y, so we can start off by subtracting 5 from both sides.

5 - y - 5 > 0 - 5

simplify.

-y > -5

divide both sides by -1.

** remember that when dividing any side by a negative, you must switch the sign! **

-y ÷ -1 < -5 ÷ -1

simplify.

y < 5

~hope i ! ~

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