Find all the complex zeros for the polynomial function. show work
f(x)=x^3-7x^2+x-7

  ±i

Step-by-step explanation:

Observing that the first two coefficients are the same as the last two, we can factor this function by grouping.

  f(x) = (x^3 -7x^2) +(x -7) = x^2(x -7) +1(x -7)

  f(x) = (x^2 +1)(x -7)

The factor x-7 has a real zero at x=7, so the complex zeros come from the quadratic factor (x^2 +1).

Setting that to zero and solving for x, we find ...

  x^2 +1 = 0

  x^2 = -1

  x = ±√(-1) = ±i

The complex zeros are x = +i and x = -i.

∠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.

its a straight line parallel to the time axis