I can't read that language but I'll guess it says write a second degree equation then solve for n when d=10.
Answer: n² - 3n - 20 = 0
That doesn't factor so there is no integer n solution.
That means there are no polygons with 10 diagonals.
The answer is z = 3 + i
z = a + bi
conj(z) = a - bi
conj(7 + 3i) = 7 - 3i
<span>(conj)z + 2z = 2 + 4i + conj(7 + 3i)
</span>a - bi + 2(a + bi) =<span> 2 + 4i + 7 - 3i
</span>a - bi + 2a + 2bi =<span> 2 + 4i + 7 - 3i
</span>3a + bi = 9 + i
From here:
3a = 9 and bi = i
a = 9/3 b = i/i
a = 3 b = 1
z = a + bi
z = 3 + 1 * i
z = 3 + i
I think it is B.
Hope its right!
:)
I believe it's r I'm really sorry if its wrong
Answer:
no
Step-by-step explanation:
y = 2x
Substitute the points into the equation
3 = 2(6)
3 = 12
This is false
The point does not make the equation true
