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The FitnessGram PACER Test is a multistage aerobic capacity test that progressively gets more difficult as it continues. The test is used to measure a student's aerobic capacity as part of the FitnessGram assessment. Students run back and forth as many times as they can, each lap signaled by a beep soundtion:

Assuming the point A has coordinates (2,3) and the center of dilation is the origin.

Then the image of point A under a dilation of scale factor of 4 is (4∗2,4∗3)=(8,12)(4*2,4*3)=(8,12)(4∗2,4∗3)=(8,12)

Hence A(2,3)→A′(8,12)A(2,3)\to A'(8,12)A(2,3)→A

′

(8,12)

In general the rule for a dilation of scale factor of k is (x,y)→(kx,ky)(x,y)\to(kx,ky)(x,y)→(kx,ky)

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∠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.

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times both sides by 2, times () by 8, add alile terms, move terms, add alike and caculate divide both sides by 3