Put the equation y = x^2 + 26 x + 160 into the form y = ( x − h )^2 + k :

Step-by-step explanation:

To write the given quadratic equation in the form  y = ( x − h )^2 + k, we need to complete the square for the given equation.

y = x^2 + 26x + 160

Shift the constant to the left side of the equation:

y - 160 = x^2 + 26x

Divide the coefficient of x by 2 and add the square of the result to both sides of the equation:

26 / 2 = 13

So adding 13^2 to both sides of the equation to get:

y - 160 + (13)^2 = x^2 + 26x + (13)^2

y - 160 + 169 = (x + 13)^2

y + 9 =  (x + 13)^2

y =  (x + 13)^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.

pretty sure 750 each month!

step-by-step explanation: