Question 3 1 pts
Tim has 74 socks, all in pairs. He has 1 more pair of white socks than black socks. How many individual socks of
each color does he have?

36 individual white socks

38 individual black socks

Step-by-step explanation:

Pairs of socks = 74

He has 1 more pair of white socks than black socks.

Take the colors of the socks to be white and black

Let

Black pairs of socks = x

White pairs of socks = x + 1

Individual socks of each color

A pair of white socks = 2 Individual white socks

= 2white

A pair of black socks = 2 individual black socks

= 2black

Total pairs of socks = black pairs of socks + white pairs of socks

74 = 2white + 2black

74 = 2(x) + 2(x +1)

74 = 2x + 2x + 2

74 = 4x + 2

Subtract 2 from both sides

74 - 2 = 4x

72 = 4x

Divide both sides by 4

x = 72/4

= 18

x = 18

Pairs of white socks = x = 18

A pair of white socks = 2whites

= 2(18)

= 36 individual white socks

Pairs of black socks = x + 1

= 18 + 1

= 19 pairs

A pair of black socks = 2black

= 2(19)

= 38 individual black socks

a box 3 3/4 feet long or 4+ feet.

step-by-step explanation:

∠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.