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3 years ago

WILL GIVE BRAINLIEST

Mathematics

2 answers:

andreyandreev [35.5K]3 years ago

7 0

Answer:

<h2>Original: M(3.5, 2) L(5.5, 2) K(7,5) J(2,5).</h2><h2>Reflected: M(3.5, -2) L(5.5, -2) K(7,-5) J(2,-5).</h2>

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).

jok3333 [9.3K]3 years ago

3 0

Answer:

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.

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allsm [11]

Answer:

See explanation

Step-by-step explanation:

16. Two parallel lines are cut by transversal. Angles with measures $(6x+20)^{\circ}$ and $(x+100)^{\circ}$ are alternate exterior angles. By alternate exterior angles, the measures of alternate exterior angles are the same:

$6x+20=x+100\\ \\6x-x=100-20\\ \\5x=80\\ \\x=16$

Then

$(6x+20)^{\circ}=(6\cdot 16+20)^{\circ}=116^{\circ}\\ \\(x+100)^{\circ}=(16+100)^{\circ}=116^{\circ}$

17. Two parallel lines are cut by transversal. Angles with measures $(2x+12)^{\circ}$ and $(3x-22)^{\circ}$ are alternate interior angles. By alternate interior angles, the measures of alternate interior angles are the same:

$2x+12=3x-22\\ \\2x-3x=-22-12\\ \\-x=-34\\ \\x=34$

Then

$(2x+12)^{\circ}=(2\cdot 34+12)^{\circ}=80^{\circ}\\ \\(3x-22)^{\circ}=(3\cdot 34-22)^{\circ}=80^{\circ}$

18. Two parallel lines are cut by transversal. Angles with measures $(6x-7)^{\circ}$ and $(5x+10)^{\circ}$ are alternate exterior angles. By alternate interior angles, the measures of alternate exterior angles are the same:

$6x-7=5x+10\\ \\6x-5x=10+7\\ \\x=17$

Then

$(6x-7)^{\circ}=(6\cdot 17-7)^{\circ}=95^{\circ}\\ \\(5x+10)^{\circ}=(5\cdot 17+10)^{\circ}=95^{\circ}$

19. The diagram shows two complementary angles with measures $2x^{\circ}$ and $56^{\circ}$ . The measures of complementary angles add up to $90^{\circ},$ then

$2x+56=90\\ \\2x=90-56\\ \\2x=34\\ \\x=17$

Hence,

$2x^{\circ}=2\cdot 17^{\circ}=34^{\circ}$

Check:

$34^{\circ}+56^{\circ}=90^{\circ}$

20. Angles $\angle 1$ and $\angle 2$ are vertical angles. By vertical angles theorem, vertical angles are congruent, so

$m\angle 1=m\angle 2\\ \\5x+7=3x+15\\ \\5x-3x=15-7\\ \\2x=8\\ \\x=4$

Hence,

$m\angle 1=(5x+7)^{\circ}=(5\cdot 4+7)^{\circ}=27^{\circ}\\ \\m\angle 2=(3x+15)^{\circ}=(3\cdot 4+15)^{\circ}=27^{\circ}$

21. $\angle 5$ and $\angle 8$ are supplementary. The measures of supplementary angles add up to $180^{\circ},$ so

$m\angle 5+m\angle 8=180^{\circ}\\ \\3x-40+7x-120=180\\ \\10x-160=180\\ \\10x=180+160\\ \\10x=340\\ \\x=34$

Therefore,

$m\angle 5=(3x-40)^{\circ}=(3\cdot 34-40)^{\circ}=62^{\circ}\\ \\m\angle 8=(7x-120)^{\circ}=(7\cdot 34-120)^{\circ}=118^{\circ}\\ \\62^{\circ}+118^{\circ}=180^{\circ}$