The area of any circle is equal to
, where r is the radius.
We know that the diameter is 12. The radius is always half of the diameter, so the radius of our circle must be 6.
Substitute 6 for r in our equation and simplify.
π6² = π6×6 =
36π in²
Alright remember, if any individual factor on the left side of the equation is equal to 0, the entire expression will be equal to 0.
k+1=0
k-5=0
Set the first factor equal to 0 and solve
k=-1
Set the next factor equal to 0 and solve
k=5
The final solution is all the values that make (k+1)(k-5)=0 true.
k=-1, 5
Hope this helped you out :)
It looks like you might have intended to say the roots are 7 + i and 5 - i, judging by the extra space between 7 and i.
The simplest polynomial with these characteristics would be
but seeing as each of the options appears to be a quartic polynomial, I suspect f(x) is also supposed to have only real coefficients. In that case, we need to pair up any complex root with its conjugate to "complete" f(x). We end up with
which appears to most closely resemble the third option. Upon expanding, we see f(x) does indeed have real coefficients:
Step-by-step explanation:
let the 2 numbers be x and x + 4 as their deference is 4.
their sum = 50
now,
→ x + x + 4 = 50
→ 2x + 4 = 50
→ 2x = 50 - 4 = 46
→ x = 46/2 = 23
therefore two numbers are,
x = 23
x + 4 = 23 + 4 = 27
hope this answer helps you dear...take care and may u have a great day ahead!
