Answer:
qn 10. 15mn² - 23m²n +4m³
Step-by-step explanation:
1. distribute 4m through the parenthesis
8mn² - 12m²n + 4m³ - 2n(5m² - 3nm) + nm(n-m)
2. use the commutative property to reorder the terms
8mn² - 12m²n + 4m³ - 2n(5m² - 3mn) + mn(n - m)
3. distribute -2n through the remaining parenthesis
8mn² - 12m²n + 4m³ -10m²n + 6mn² + mn² - m²n
4. collect like terms
8mn² + 6mn² + mn² - 12m²n - 10m²n - m²n + 4m³
5. complete bodmas
15mn² - 23m²n +4m³
that's is how you do it so the answer is
15mn² -23m²n + 4m³
Answer:
attached below
Step-by-step explanation:
Attached below is a detailed solution to your problem above
a) show that the maximum likelihood estimator of Ï2 is Xi and Yi.
attached below is the detailed solution
Answer:
The radian measure of the angle drawn in standard position that corresponds with the ray containing the coordinate point
is approximately
radians.
Step-by-step explanation:
With respect to origin, the coordinate point belongs to the third quadrant, which comprises the family of angles from
to
. The angle in standard position can be estimated by using the following equivalence:
![\theta = \pi\,rad + \tan^{-1} \left(\frac{3\sqrt{2}}{12} \right)](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%5Cpi%5C%2Crad%20%2B%20%5Ctan%5E%7B-1%7D%20%5Cleft%28%5Cfrac%7B3%5Csqrt%7B2%7D%7D%7B12%7D%20%5Cright%29)
![\theta \approx \pi \,rad + 0.108\pi \,rad](https://tex.z-dn.net/?f=%5Ctheta%20%5Capprox%20%5Cpi%20%5C%2Crad%20%2B%200.108%5Cpi%20%5C%2Crad)
![\theta \approx 1.108\pi\,rad](https://tex.z-dn.net/?f=%5Ctheta%20%5Capprox%201.108%5Cpi%5C%2Crad)
The radian measure of the angle drawn in standard position that corresponds with the ray containing the coordinate point
is approximately
radians.
It is equal to 2(x)^3 because pemdas, exponnets come before multiplication so it would be x^3 then multiply whatever that is by 2
2x^3 is answer (4th choice)