A box is dragged across 202020 meters with a force of 606060 Newtons, which are\dfrac{\text{kg}\cdot\text{m}}{\ text{s}^2} s 2 kg⋅m start fraction, start text, k, g, end text, dot, start text, m, end text, divided by, start text, s, end text, squared, end fraction. What calculation will give us the work done, in Joules (which are \dfrac{\text{kg}\cdot\text{m}^2}{\t ext{s}^2} s 2 kg⋅m 2 start fraction, start text, k, g, end text, dot, start text, m, end text, squared, divided by, start text, s, end text, squared, end fraction)? Choose 1

∠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.

d i think

step-by-step explanation:

8(4)^8=72^feet to say that the total can conclude to 8(4) which as you know would mean 8 x 4. that equals 32w. 32^ could make a 72 answer as well. but your coefficent will remain w so your answer is 72w.

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write out the polynomial, with leading coefficient 1:

y = 1 (x+2)^1 (x-7)^1 (x-5)^2

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