Which operation is performed in the derivation of the

I will give a brainlest, i promise, please help me with this.

valkas [14]

Answer:

1) b and m

2) m∠8=m∠6

3) 160°

4)x=60°

Step-by-step explanation:

1) all straight lines sum to 180°

subtract the angles given from 180°

the other angle for b is 25°, while the other angle for m is 155°

so we can see that the angles for both lines are the same, hence they are parallel.

2) ∠8 should be the same as ∠6, ∠10 should be the same as ∠3, ∠7 same as ∠5 and ∠9 same as ∠4

in the options we are only given '∠8 should be the same as ∠6' as the correct answer, so we take that.

3) from the image we can see that both horizontal lines are parallel to each other, so both angles on the lines should be same, so ∠CET would be (2x-16)°

(2x-16)°+(7x+20)°=180°

we get x=20(nearest whole number)

∠CED=7x+20=7(20)+20=160°

4) since we need to show that they are parallel,

(2x+30)°=(4x-90)°

2x-4x=-90-30

-2x=-120

x=60

we then plug the x value into the two equations, in which we get 150° for both the angles [2(60)+30=4(60)-90] ⇒ (150=150)

I hope u understand it the way I put it.

8 0

2 years ago

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Ad libitum [116K]

Answer:

(x + 6, y + 0), 180° rotation, reflection over the x‐axis

Step-by-step explanation:

The answer can be found out simply , a trapezoid has its horizontal sides usually parallel meanwhile the vertical sides are not parallel.

The horizontal parallel sides are on the x-axis.

Reflection over y- axis would leave the trapezoid in a vertical position such that the trapezoid ABCD won't be carried on the transformed trapezoid as shown in figure.

So option 1 and 2 are removed.

Now, a 90 degree rotation would leave the trapezoid in a vertical position again so its not suitable again.

In,The final option (x + 6, y + 0), 180° rotation, reflection over the x‐axis, x+6 would allow the parallel sides to increase in value hence the trapezoid would increase in size,

180 degree rotation would leave the trapezoid in an opposite position and reflection over x-axis would bring it below the Original trapezoid. Hence, transformed trapezoid A`B`C`D` would carry original trapezoid ABCD onto itself