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It becomes: 2k squared
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What does odd X even equal?
Let even be of the form 2k and odd be of the form 2k+1. Then odd * even becomes 2k*2k+1, or 4k^2 +2k. This can be written as 2(k^2 + k), which is of the form 2k. Therefore, odd X even equals even.
How do you find the straight line equation given only that the slope is k passing through a point 3 and 2k on the line?
Suppose P = (x, y) are the coordinates of any point on the line. Then the segment of the line joining P to the point (3, 2k) has slope k That is, (y - 2k) / (x - 3) = k Simplifying, y - 2k = kx - 3k or y = kx - k equivalently, y = k(x - 1)
-3k plus 10 equals k plus 2?
-4. -3k+10=k+2.... First we have to move the K to one side -3k+10=k+2 __________ This will reduce -3k to -2k and k+2 to just 2 K -2k+10=2 Now we have to subtract 10 to both sides as to get k alone. -2k+10=2 ________ Now we should have -2k on one side and 2-10 on the other. -10 -2k=2-10....... 2-10= 8. We have to get -2k into just K on itself so we divide by 2. -2k=-8 ______ Your answer is 4. You have to do one side what you do the other. 2
2k plus 3 equals 11 what is k?
2k + 3 = 11(- 3 from both sides)2k = 8(divide both sides by 2)k = 4
K - 20 equals 2k - 13?
k=-7
What does odd X even equal?
Let even be of the form 2k and odd be of the form 2k+1. Then odd * even becomes 2k*2k+1, or 4k^2 +2k. This can be written as 2(k^2 + k), which is of the form 2k. Therefore, odd X even equals even.
How do you find the straight line equation given only that the slope is k passing through a point 3 and 2k on the line?
Suppose P = (x, y) are the coordinates of any point on the line. Then the segment of the line joining P to the point (3, 2k) has slope k That is, (y - 2k) / (x - 3) = k Simplifying, y - 2k = kx - 3k or y = kx - k equivalently, y = k(x - 1)
9 minus 2k-3 equals k show work and solve?
9 - 2k - 3 = k Add 2k to both sides: 9 - 3 = k + 2k Combine like terms: 6 = 3k Divide both sides by 3: 2 = k
-3k plus 10 equals k plus 2?
-4. -3k+10=k+2.... First we have to move the K to one side -3k+10=k+2 __________ This will reduce -3k to -2k and k+2 to just 2 K -2k+10=2 Now we have to subtract 10 to both sides as to get k alone. -2k+10=2 ________ Now we should have -2k on one side and 2-10 on the other. -10 -2k=2-10....... 2-10= 8. We have to get -2k into just K on itself so we divide by 2. -2k=-8 ______ Your answer is 4. You have to do one side what you do the other. 2
2k plus 3 equals 11 what is k?
2k + 3 = 11(- 3 from both sides)2k = 8(divide both sides by 2)k = 4
K - 20 equals 2k - 13?
k=-7
If 2k-1 equals 0 then k equals?
2k - 1 = 0 Add 1 to both sides: 2k = 1 Divide both sides by two: k = 0.5
How do you factor -2k - k3 - 3k2?
-(k(k + 2)(k + 1))
How do you find the LCM of k to the 2nd power k to the 2nd power-1 and k to the 2nd power minus 2k plus 1?
Factor them. k2 = k x k k2 - 1 = (k - 1)(k + 1) k2 - 2k + 1 = (k - 1)(k - 1) Combine the factors, eliminating duplicates. k2(k + 1)(k - 1)(k - 1) = k5 - k4 - k3 + k2, the LCM
When cosec is positive?
Between (2k)*pi radians and (1+2k)*pi radians where k is an integer. If you are still working with degrees, that is360*k degrees to (1+2k)*180 degrees, for integer values of k.NB: these are open intervals: that is, the end points are not included.Between (2k)*pi radians and (1+2k)*pi radians where k is an integer. If you are still working with degrees, that is360*k degrees to (1+2k)*180 degrees, for integer values of k.NB: these are open intervals: that is, the end points are not included.Between (2k)*pi radians and (1+2k)*pi radians where k is an integer. If you are still working with degrees, that is360*k degrees to (1+2k)*180 degrees, for integer values of k.NB: these are open intervals: that is, the end points are not included.Between (2k)*pi radians and (1+2k)*pi radians where k is an integer. If you are still working with degrees, that is360*k degrees to (1+2k)*180 degrees, for integer values of k.NB: these are open intervals: that is, the end points are not included.
What is 2k-3k-m plus 2m?
2k - 3k - m + 2m = -k + m
You is the abbreviation for?
"K" k stands for thousand like 2k for the year 2000
