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Since x2 - 1 is a multiple of x - 1, it is the LCM of this problem.

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Q: What is the least common multiple of x2-1 and x-1?
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Is the smallest LCM of a number is the number itself?

Umm, you're question seems to have been a bit garbled. If the question was meant to be: Is the LCM of a single number the number itself? In that case the answer is...N/A... LCM (Least Common Multiple) MUST be at least two numbers. Otherwise "common" has no meaning in the title and the least multiple of any number would be 1 x0 =1, x1 =x, etc...However, if the question was meant to be something like: If a number is a multiple of another number, is their LCM the larger number? In that case, yes.P.S. "Is the smallest lcm" is redundant. Smallest and least are identical in this situation.


How do you use a variance-covariance matrix to obtain least squares estimates?

Suppose that you have simple two variable model: Y=b0+b1X1+e The least squares estimator for the slope coefficient, b1 can be obtained with b1=cov(X1,Y)/var(X1) the intercept term can be calculated from the means of X1 and Y b0=mean(Y)-b1*mean(X1) In a larger model, Y=b0+b1X1+b2X2+e the estimator for b1 can be found with b1=(cov(X1,Y)var(X2)-cov(X2,Y)cov(X1,X2))/(var(X1)var(X2)-cov(X1,X2)2) to find b2, simply swap the X1 and X2 terms in the above to get b2=(cov(X2,Y)var(X1)-cov(X1,Y)cov(X1,X2))/(var(X1)var(X2)-cov(X1,X2)2) Find the intercept with b0=mean(Y)-b1*mean(X1)-b2*mean(X2) Beyond two regressors, it just gets ugly.


How matrix X1 X2 equals matrix X1 X2?

i think its pretty much the same thing because matrix X1 X2 IS ACTUALLY X1 X2


How do you do equations and graphs for slope?

The equation for the slope between the points A = (x1, y1) and B = (x2, y2) = (y2 - y1)/(x2 - x1), provided x1 is different from x2. If x1 and x2 are the same then the slope is not defined.


What is the slope formula in equation form?

If (x1, y1) and (x2, y2) are two points on the line, then the formula for the slope is (y2-y1)/(x2-x1) provided x2 ≠ x1. If x2 = x1 then the line is vertical and the slope is not defined.

Related questions

What is the least common multiple of 3x4 and 12x2?

Not sure this question is stated correctly. Answering logically, the LCM of 12 (3x4) and 24 (12x2) is 24. This is because 12(x2=24) and 24(x1=24) are both factors of 24. Altering the question slightly, the least common multiple of 2, 3, 4, and 12 is 12. This is because 2(x6=12), 3(x4=12), 4(x3=12), and 12(x1=12) are all factors of 12.


Is the smallest LCM of a number is the number itself?

Umm, you're question seems to have been a bit garbled. If the question was meant to be: Is the LCM of a single number the number itself? In that case the answer is...N/A... LCM (Least Common Multiple) MUST be at least two numbers. Otherwise "common" has no meaning in the title and the least multiple of any number would be 1 x0 =1, x1 =x, etc...However, if the question was meant to be something like: If a number is a multiple of another number, is their LCM the larger number? In that case, yes.P.S. "Is the smallest lcm" is redundant. Smallest and least are identical in this situation.


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How do you use a variance-covariance matrix to obtain least squares estimates?

Suppose that you have simple two variable model: Y=b0+b1X1+e The least squares estimator for the slope coefficient, b1 can be obtained with b1=cov(X1,Y)/var(X1) the intercept term can be calculated from the means of X1 and Y b0=mean(Y)-b1*mean(X1) In a larger model, Y=b0+b1X1+b2X2+e the estimator for b1 can be found with b1=(cov(X1,Y)var(X2)-cov(X2,Y)cov(X1,X2))/(var(X1)var(X2)-cov(X1,X2)2) to find b2, simply swap the X1 and X2 terms in the above to get b2=(cov(X2,Y)var(X1)-cov(X1,Y)cov(X1,X2))/(var(X1)var(X2)-cov(X1,X2)2) Find the intercept with b0=mean(Y)-b1*mean(X1)-b2*mean(X2) Beyond two regressors, it just gets ugly.


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