Are there any values of x for which the areas of the figures are the same? Explain. Set the area of the green triangle equal to the area of the blue rectangle.

Area of green triangle=Area of blue rectangle
=

Question 2
There
value(s) of x for which the areas of the figures are the same, because there
solution(s) to the equation.

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.

sample answer: you can determine whether each element of the domain is paired with exactly one element of the range. for example, if given a graph, you could use the vertical line test; if a vertical line intersects the graph more than once, then the relation that the graph represents is not a function.

the correct answer would be the third option

arc length = radius * x ( in radians)

6 * pi = 18 * x

divide by 18 on each side

6 pi / 18 = x

pi/3 = x

the angle is pi / 3 radians

or   60 degrees