In a regular octagon it is impossible to have an acute angle!
Hope this helps!!
Answer:
Step-by-step explanation:
We want to find a third degree polynomial with zeros <em>x </em>= 2 and <em>x</em> = 2i and f(-1) = 30.
First, note that by the Complex Root Theorem, since <em>x</em> = 2i is a root, <em>x</em> = -2i must also be a root.
Hence, we will have the three factors:
Where <em>a</em> is the leading coefficient.
Expand and simplify the second and third factors:
Hence:
Since f(-1) = 30:
In conclusion, third degree polynomial function is:
Answer:
C
Step-by-step explanation:
:)
Answer:
The pairs are (13,15) and (-15,-13).
Step-by-step explanation:
If n is an odd integer, the very next odd integer will be n+2.
n+1 is even (so we aren't using this number)
The sum of the squares of (n) and (n+2) is 394.
This means
(n)^2+(n+2)^2=394
n^2+(n+2)(n+2)=394
n^2+n^2+4n+4=394 since (a+b)(a+b)=a^2+2ab+b^2
Combine like terms:
2n^2+4n+4=394
Subtract 394 on both sides:
2n^2+4n-390=0
Divide both sides by 2:
n^2+2n-195=0
Now we need to find two numbers that multiply to be -195 and add up to be 2.
15 and -13 since 15(-13)=-195 and 15+(-13)=2
So the factored form is
(n+15)(n-13)=0
This means we have n+15=0 and n-13=0 to solve.
n+15=0
Subtract 15 on both sides:
n=-15
n-13=0
Add 13 on both sides:
n=13
So if n=13 , then n+2=15.
If n=-15, then n+2=-13.
Let's check both results
(n,n+2)=(13,15)
13^2+15^2=169+225=394. So (13,15) looks good!
(n,n+2)=(-15,-13)
(-15)^2+(-13)^2=225+169=394. So (-15,-13) looks good!
