Abigail and Spencer are correct, Lauren is incorrect.
If the circle is at the origin and (0,4) is a point on the circle then the radius of the circle is:
r^2=0^2+4^2
r^2=16
r=4
so we can check that the other point is four units from the origin and thus also on the circle...
d^2=2^2+6^2
d^2=4+36
d^2=40
since d^2>r^2, (2,6) is outside the circle...
Answer:
b.-5
Step-by-step explanation:
Whatever you do to one side you do to the other
D because u have to look the fraction
Roots with imaginary parts always occur in conjugate pairs. Three of the four roots are known and they are all real, which means the fourth root must also be real.
Because we know 3 and -1 (multiplicity 2) are both roots, the last root
is such that we can write

There are a few ways we can go about finding
, but the easiest way would be to consider only the constant term in the expansion of the right hand side. We don't have to actually compute the expansion, because we know by properties of multiplication that the constant term will be
.
Meanwhile, on the left hand side, we see the constant term is supposed to be 9, which means we have

so the missing root is 3.
Other things we could have tried that spring to mind:
- three rounds of division, dividing the quartic polynomial by
, then by
twice, and noting that the remainder upon each division should be 0
- rational root theorem