Directed line segment overline FG has endpoints F(4, 9) and G(- 8, - 9) .On the coordinate plane below , plot a point that partitions the directed line segment in a ratio of Point 1/2 Undo Reda x Reset 10 8 7 65 443 2

The coordinates of the point that makes the division in the given ratio is (0,3)

Step-by-step explanation:

Here, we want to find the point on the line segment that divides the line segment in the ratio 1:2

We simply use the internal division formula

That would be;

(x,y) = (nx1 + mx2)/(m + n) , (ny1 + my2)/(m + n)

m = 1 and n = 2

(x1,y1) = (4,9)

(x2,y2) = (-8,-9)

Substituting these values into the formula, we have;

2(4) + 1(-8)/(1 + 2) , 2(9) + 1(-9)/(2 + 1)

= (8-8)/3 , (18-9)/3

= (0,3)

4+n=5

step-by-step explanation:

this is because the four is four hours and the n represents the 1 extra hour over the weekend.

b= i'm srry i don't know that one.

4+n=5 n=1 because if vernon plays 5 hours a week then 4+1=5 hours.

hope this : )

∠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.