(4y^2-3y+2)+(5y+12y-4)-(13y^2-6y+20 )

-(9 y^2 - 20 y + 22) or if you need y = 10/9 + (7 i sqrt(2))/9 or y = 10/9 - (7 i sqrt(2))/9

Step-by-step explanation:

Simplify the following:

-(13 y^2 - 6 y + 20) + 4 y^2 + 12 y + 5 y - 3 y - 4 + 2

Grouping like terms, -(13 y^2 - 6 y + 20) + 4 y^2 + 12 y + 5 y - 3 y - 4 + 2 = -(13 y^2 - 6 y + 20) + 4 y^2 + (-3 y + 5 y + 12 y) + (2 - 4):

-(13 y^2 - 6 y + 20) + 4 y^2 + (-3 y + 5 y + 12 y) + (2 - 4)

-3 y + 5 y + 12 y = 14 y:

-(13 y^2 - 6 y + 20) + 4 y^2 + 14 y + (2 - 4)

2 - 4 = -2:

-(13 y^2 - 6 y + 20) + 4 y^2 + 14 y + -2

-(13 y^2 - 6 y + 20) = -13 y^2 + 6 y - 20:

-13 y^2 + 6 y - 20 + 4 y^2 + 14 y - 2

Grouping like terms, 4 y^2 - 13 y^2 + 14 y + 6 y - 20 - 2 = (-13 y^2 + 4 y^2) + (6 y + 14 y) + (-20 - 2):

(-13 y^2 + 4 y^2) + (6 y + 14 y) + (-20 - 2)

4 y^2 - 13 y^2 = -9 y^2:

-9 y^2 + (6 y + 14 y) + (-20 - 2)

6 y + 14 y = 20 y:

-9 y^2 + 20 y + (-20 - 2)

-20 - 2 = -22:

-9 y^2 + 20 y + -22

Factor -1 out of -9 y^2 + 20 y - 22:

-(9 y^2 - 20 y + 22)

Solve for y:

-9 y^2 + 20 y - 22 = 0

Divide both sides by -9:

y^2 - (20 y)/9 + 22/9 = 0

Subtract 22/9 from both sides:

y^2 - (20 y)/9 = -22/9

Add 100/81 to both sides:

y^2 - (20 y)/9 + 100/81 = -98/81

Write the left hand side as a square:

(y - 10/9)^2 = -98/81

Take the square root of both sides:

y - 10/9 = (7 i sqrt(2))/9 or y - 10/9 = -(7 i sqrt(2))/9

Add 10/9 to both sides:

y = 10/9 + (7 i sqrt(2))/9 or y - 10/9 = -(7 i sqrt(2))/9

Add 10/9 to both sides:

y = 10/9 + (7 i sqrt(2))/9 or y = 10/9 - (7 i sqrt(2))/9

∠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.

answer: my brother can figure it out, imma go get him so i can answer : ) edit: my brother is sleeping, sorry man

step-by-step explanation: