For any nonempty set $t$ whose elements are positive integers, define $f(t)$ to be the square of the product of the elements of $t$. for example, if $t=\{1,3,6\}$, then $f(t)=(1\cdot 3\cdot 6)^2 = 18^2 = 324$. consider the nonempty subsets $t$ of $\{1,2,3,4,5,6,7\}$ that do not contain two consecutive integers. if we compute $f(t)$ for each such set, then add up the resulting values, what do we get?

We get

$225+324+441+576+784+1125+2304+3136+4900+11025=24840$

Step-by-step explanation:

$(1\cdot 3\cdot 5)^2 = 15^2 = 225$

$(1\cdot 3\cdot 6)^2 = 18^2 = 324$

$(1\cdot 3\cdot 7)^2 = 21^2 = 441$

$(1\cdot 4\cdot 6)^2 = 24^2 = 576$

$(1\cdot 4\cdot 7)^2 = 28^2 = 784$

$(1\cdot 5\cdot 7)^2 = 35^2 = 1125$

$(2\cdot 4\cdot 6)^2 = 48^2 = 2304$

$(2\cdot 7\cdot 7)^2 = 56^2 = 3136$

$(2\cdot 5\cdot 7)^2 = 70^2 = 4900$

$(3\cdot 5\cdot 7)^2 = 105^2 = 11025$

We get

$225+324+441+576+784+1125+2304+3136+4900+11025=24840$

∠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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