16w^2+18w+1

I need the you to show work

w=(-9+√65)/16 or ((-9-√65)16

Step-by-step explanation:

This quadratic cannot be factorises using the method of factorization so for this example I choose to use the general quadratic formula;

x=(-b±√b²-4ac)/2a

By comparing 16w²+18w+1 with ax²+bx+c, a=16,b=18,c=1 and we plug them into the formula and simplify

w={-18±√(18)²-4(16×1)}/2(16)

w={-18±√324-64}/32

w={-18±√260}/32

w=(-18±2√65)/32

w=(-9+√65)/16 or ((-9-√65)16

∠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.

the answer is two. hope this !