PLZ help now i need help n

7

Step-by-step explanation:

hey there,

< Here's how you do subtraction in fractions!

Make sure you find a least common denominator. Here, it's given to you to be 15. Usually, you can just multiply the two denominators but then it would be quite large at times, so you kind of just have to guess.

How did they go from    to ?

Well, they used a least common denominator. 5 was multiplied by 3 to create 15, and 3 was multiplied by 5. But when you multiply the bottoms, you have to multiply the tops of the fraction too (the numerators).

So 4/5x was multiplied by 3. Denominator turned out to be 15, and now the top is 4x3 = 12. So the fraction is 12/15.

1/3x, on the other hand, was multiplied by 5. Denominator was also 15 so they're the same, and top is 1x5 = 5. So the fraction is 5/15.

Now, let's put them together.

When subtracting x, the x just stays there, so if it was 10x-4x, the answer would've been 6x. So now let's subtract!

12-5 = 7, and the denominator stays the same.

The answer is . So the numerator, or answer to your question, would be 7.

Hope this helped! Feel free to ask anything else.

∠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.