1. y varies inversely with x. if y = 3 and k (the constant of variation) = 33, what is x?

round to the nearest whole number, if necessary.

2. y varies inversely with x, and k (the constant of variation) = 5.6. what is the value of y when x = 4?

round to the nearest tenth, if necessary.

3. y varies inversely with x. if y = 17 and k (the constant of variation) = 76, what is x?

round to the nearest tenth, if necessary.

Step-by-step explanation:

1) y varies inversely with x. If y = 3 and k (the constant of variation) = 33

If y varies inversely with x, then as y increases increases, x reduces and as y reduces, x increases.

y varies inversely with x

Introducing constant of inverse variation, k

y = k/x

If y = 3 and k (the constant of variation) = 33,

x = k / y = 33/3

= 11

2) y varies inversely with x, and k (the constant of variation) = 5.6. What is the value of y when x = 4

y = k/x

y = 5.6/4 = 1.4

3) y varies inversely with x. If y = 17 and k (the constant of variation) = 76

y = k /x

x = k /y = 76/17 = 4.4705

= 4.5 to the nearest tenth

pooopoopopoopop

step-by-step explanati∵∵∴∴on:

∠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.