Ali has a collection of 177 snap-together building blocks. Ali has one fourth as many red blocks as blue blocks, half as many white blocks as black blocks, an equal number of black and blue blocks, and 34 miscellaneous blocks of other colors. How many red blocks does Ali have? Help me plz!

The number of red blocks Ali has is 13 red balls

Step-by-step explanation:

The given parameters are;

The number of snap-together building blocks Ali has = 177

The number of red blocks Ali has = 1/4 × The number of blue blocks Ali has

The number of white  blocks Ali has = 1/2 × The number of black blocks Ali has

The number of black blocks Ali has = The number of blue blocks he has

The number of miscellaneous blocks Ali has = 34

Let x represent the number of blue blocks Ali has, therefore, we have;

The number of red blocks Ali has = 1/4 × x

The number of black blocks Ali has = The number of blue blocks = x

x = 1/2 × The number of white blocks Ali has

∴ The number of white blocks Ali has = 2·x

The number of known colored blocks Ali has = 177 - 34 = 143

Therefore, we have;

x + 1/4·x + x + 1/2·x = 143

11/4·x = 143

x = 143 × 4/11 = 52

Therefore, the number of blue blocks Ali has = x = 52 blue balls

The number of red blocks Ali has = 1/4 × The number of blue blocks Ali has = 1/4 × 52 = 13

The number of red blocks Ali has = 13 red balls.

the one that has a slope of 1/2 and is positive.

step-by-step explanation: 2 buckets 1 trip ; 4 buckets 2 trips and so 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.