Will give brainliest
(this is a long answer)
if trapezoid jklm is reflected over the x-axis, what do you notice about the ordered pairs of the original figure and the ordered pairs of its reflection over the x-axis?

J'(-2,5)

K'(-7,5)

M'(-3.5,2)

L'(5.5,2)

Step-by-step explanation:

Please refer to the image attached to this answer.

Trapezoid JKLM is reflected over the x-axis, what do we notice about the ordered pairs of the original figure and the ordered pairs of its reflection over the x-axis is described below.

Point J , K L and M  , when reflected over x axis , there distance from the y axis remains the same in second quadrant. the x coordinate becomes negative however  the y coordinates remains the same.

The coordinates will be as shown in the image attached.

Step-by-step explanation:

The coordinates of the given trapezoid JKLM are

M(3.5, 2) L(5.5, 2) K(7,5) J(2,5).

If we want to reflect the trapezoid JKLM over the x-axis, we just need to change its vertical coordinates for their opposite, that is

M(3.5, -2) L(5.5, -2) K(7,-5) J(2,-5).

There reason of this is because if we need a reflection across x-axis, what changes is vertical coordintes, because the figure must go from I quadrant to IV quadrant where y-valus are negative, and x-values remains positive.

Therefore, the ordered pair of the original figure are:

M(3.5, 2) L(5.5, 2) K(7,5) J(2,5).

The ordered pair of the reflection are:

M(3.5, -2) L(5.5, -2) K(7,-5) J(2,-5).

step-by-step explanation:

through just a quick inspection, one can see that the graph of this cubing function starts in quadrant iii and ends in quadrant i.   when x = 0, y = 3, so the y-intercept is (0, 3).

if graphs are part of the problem statement, share them.   .

find the magnitude of the trig term  

csc

(

x

)

−

1

by taking the absolute value of the coefficient.

1

find the lower bound of the range.

tap for more

y

=

−

2

find the upper bound of the range.

tap for more

y

=

0

the range is  

y

≤

−

2

or  

y

≥

0

.

(

−

∞

,

−

2

]

∪

[

0

,

∞

)

{

y

|

y

≤

−

2

,

y

≥

0

}

step-by-step explanation: