shut up and do your hw
i think its pretty much the same thing because matrix X1 X2 IS ACTUALLY X1 X2
4
It is linear. The highest power is 1 (x = x1, y = y1) so it is linear.
This question can only be answered if the probability distribution functions of X1, X2 and X3 are known. They are not and so the question cannot be answered.
it equals x1 it equals x1
3
shut up and do your hw
Assuming you mean x=1, then it is a solution.
f(x1) = (-5)2 + 3*(-5) + 5.1 = 25 - 15 + 5.1 = 15.1 f(x2) = (6)2 + 3*(6) + 5.1 = 36 + 18 + 5.1 = 59.1 f(x2)-f(x1) = 59.1 - 15.1 = 44 x2 - x1 = 6 - (-5) = 11 So average rate of change = 44/11 = 4
x²+6x+9=49 x²+6x-40=0 x1=-6/2 - Square root of ((6/2)²+40) x1=-3 - 7 x1= -10 x2=-6/2 + Square root of ((6/2)²+40) x2=-3 + 7 x2= 4
i think its pretty much the same thing because matrix X1 X2 IS ACTUALLY X1 X2
4
It is linear. The highest power is 1 (x = x1, y = y1) so it is linear.
Given the limited information in the question, Z is maximised when x1 or x2 (or both) are maximised. There is no trade-off between x1 and x2 to worry about.
This question can only be answered if the probability distribution functions of X1, X2 and X3 are known. They are not and so the question cannot be answered.
minimize Z=-X1+2X2=