P(A) = 1/4
p(B) = 2/3
p(A ∩ B) = 1/5
p(A).p(B) = 2/12 = 1/6 ≠p(A ∩ B )
thus no, they are dependent because P(A) ⋅ P(B) ≠ P(A ∩ B)
Answer:
![(2-5i)(p+q)(i)=29i](https://tex.z-dn.net/?f=%282-5i%29%28p%2Bq%29%28i%29%3D29i)
Step-by-step explanation:
We have the product of 2 complex numbers
![(2-5i)(p+q)(i)](https://tex.z-dn.net/?f=%282-5i%29%28p%2Bq%29%28i%29)
We know that:
![p=2\\\\q=5i](https://tex.z-dn.net/?f=p%3D2%5C%5C%5C%5Cq%3D5i)
Then we substitute these values in the expression
![(2-5i)((2)+(5i))(i)](https://tex.z-dn.net/?f=%282-5i%29%28%282%29%2B%285i%29%29%28i%29)
![(2-5i)(2+5i)(i)](https://tex.z-dn.net/?f=%282-5i%29%282%2B5i%29%28i%29)
The product of a complex number
by its conjugate
is always equal to:
![a ^ 2 - (bi) ^ 2](https://tex.z-dn.net/?f=a%20%5E%202%20-%20%28bi%29%20%5E%202)
Then
![(2-5i)(2+5i)(i)=(2^2-5^2i^2)(i)](https://tex.z-dn.net/?f=%282-5i%29%282%2B5i%29%28i%29%3D%282%5E2-5%5E2i%5E2%29%28i%29)
Remember that:
![i=\sqrt{-1}\\\\i^2 = -1](https://tex.z-dn.net/?f=i%3D%5Csqrt%7B-1%7D%5C%5C%5C%5Ci%5E2%20%3D%20-1)
So
![(2^2-5^2i^2)(i)= (4 - 25(-1))(i)\\\\(4 - 25(-1))(i) = (4+25)i=29i](https://tex.z-dn.net/?f=%282%5E2-5%5E2i%5E2%29%28i%29%3D%20%284%20-%2025%28-1%29%29%28i%29%5C%5C%5C%5C%284%20-%2025%28-1%29%29%28i%29%20%3D%20%284%2B25%29i%3D29i)
Finally
![(2-5i)(p+q)(i)=29i](https://tex.z-dn.net/?f=%282-5i%29%28p%2Bq%29%28i%29%3D29i)
Answer:
The correct solution is 4
Step-by-step explanation:
64÷4^2
64÷16
=4
(2x + 10)(x - 9)
2x(x - 9) + 10(x - 9)
2x(x) - 2x(9) + 10(x) - 10(9)
2x² - 18x + 10x - 90
2x² - 8x - 90