Answer and Step-by-step explanation:
Polynomial models are an excellent implementation for determining which input element reaction and their direction. These are also the most common models used for the scanning of designed experiments. It defines as:
Z = a0 + a1x1 + a2x2 + a11x12 + a22x22+ a12x1x2 + Є
It is a quadratic (second-order) polynomial model for two variables.
The single x terms are the main effect. The squared terms are quadratic effects. These are used to model curvature in the response surface. The product terms are used to model the interaction between explanatory variables where Є is an unobserved random error.
A polynomial term, quadratic or cubic, turns the linear regression model into a curve. Because x is squared or cubed, but the beta coefficient is a linear model.
In general, we can model the expected value of y as nth order polynomial, the general polynomial model is:
Y = B0 + B1x1 + B2x2 + B3x3 + … +
These models are all linear since the function is linear in terms of the new perimeter. Therefore least-squares analysis, polynomial regression can be addressed entirely using multiple regression
Answer:
i think it would be 21 percent he got wrong so if he got 21 percent wrong it would be 79 percent right
Step-by-step explanation:
beacause if it was 10 question it would be 30 percent wrong since its 15 you divide it by 100 which would be 7 and 7 times 3 is 21
sorry if im wrong
Well that was nice of him
Answer: 10*(cos(pi) + i*sin(pi))
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Work Shown:
z1 = 1+i is in the form a+bi where a = 1 and b = 1
r = sqrt(a^2+b^2)
r = sqrt(1^2+1^2)
r = sqrt(2)
theta = arctan(b/a)
theta = arctan(1/1)
theta = pi/4
The polar form for z1 is
z1 = r*(cos(theta) + i*sin(theta))
z1 = sqrt(2)*(cos(pi/4)+i*sin(pi/4)
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z2 = -5+5i is in the form a+bi where a = -5 and b = 5
r = sqrt(a^2+b^2)
r = sqrt((-5)^2+5^2)
r = sqrt(50)
r = 5*sqrt(2)
theta = arctan(b/a)
theta = arctan(5/(-5))
theta = 3pi/4
The polar form for z2 is
z2 = r*(cos(theta) + i*sin(theta))
z2 = 5*sqrt(2)*(cos(3pi/4)+i*sin(3pi/4)
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Now use the rule
If
z = a*(cos(b) + i*sin(b)) and w = c*(cos(d)+i*sin(d))
then
z*w = a*c*(cos(b+d)+i*sin(b+d))
We have
a = sqrt(2)
b = pi/4
c = 5*sqrt(2)
d = 3pi/4
So...
z*w = a*c*(cos(b+d)+i*sin(b+d))
z1*z2 = sqrt(2)*5*sqrt(2)*(cos(pi/4+3pi/4)+i*sin(pi/4+3pi/4))
z1*z2 = 10*(cos(4pi/4)+i*sin(4pi/4))
z1*z2 = 10*(cos(pi)+i*sin(pi))
which is the answer in polar form
