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3k + 3 = 8 3k = 8 -3 3k = 5 k = 5/3
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)
2k + 3 = 11(- 3 from both sides)2k = 8(divide both sides by 2)k = 4
2k = 5k-30 Subtract 5k from both sides: -3k = -30 Divide both sides by -3 to find the value of k remembering that a minus divided into a minus becomes a plus: k = 10
-47 + 6k = 3k - 2Add 47 to each side:6k = 3k +45Subtract 3k from each side:3k = 45Divide each side by 3:k = 15
By using algebra.
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-1=k+2 2k=3 k=3/2=1.5
The question is unclear, so the author will provide answers for a number of interpretations: 1. 3k-6(2k+1) = 3k-12k-6=-9k-6=-3(3k+2) 2. 3k-6(2k)+1=3k-12k+1=-9k+1 3. (3k-6)(2k)+1 = 6k^2 -12k + 1 = 6(k-1-sqrt(5/6))(k-1+sqrt(5/6)) 4. (3k-6)(2k+1) = 6k^2 - 12k + 3k - 6 = 6k^2 -9k + 6 = 3(2k^2 - 3k + 2) Line 4 cannot be factorised further. sqrt and ^2 refer to the square root, and squared respectively. Lines 1 and 2 require knowledge of expansion of linear equations, addition of like terms, and factorisation of linear equations. Lines 3 and 4 also require knowledge of addition of like terms, and expansion and factorisation of quadratic equations. In no case can an exact value for k be determined as we were given an expression rather than an equality.
3k + 3 = 8 3k = 8 -3 3k = 5 k = 5/3
6k2 - k - 12 (2k - 3)(3k + 4) k = 3/2 and -4/3 1.5 and -1.33 are the factors
1). Add ' 3k ' to each side of the equation. 2). Add ' 5 ' to each side. 3). Divide each side by ' 3 ' .
All even numbers greater than 2 are composite because they are all divisible by 2. Therefore, from 3 onwards, all prime numbers are odd. Consider three consecutive odd numbers. They must be of the form 2n+1, 2n+3 and 2n+5 where n is an integer. Since n is an integer, n will leave a remainder of 0, 1 or 2 when it is divided by 3. Suppose n leaves a remainder of 0 when divided by 3. Therefore n = 3k for some integer k. Then 2n+3 = 2*(3k) + 3 = 6k + 3 = 3*(2k+1). That is, middle of the three consecutive odd numbers is divisible by 3 and so it is not a prime. Now, suppose n leaves a remainder of 1 when divided by 3. Therefore n = 3k+1 for some integer k. Then 2n+1 = 2*(3k+1) + 1 = 6k+2+1 = 6k+3 = 3*(2k+1). That is, first of the three consecutive odd numbers is divisible by 3 and so it is not a prime. Finally, suppose n leaves a remainder of 2 when divided by 3. Therefore n = 3k+2 for some integer k. Then 2n+5 = 2*(3k+2) + 5 = 6k+4+5 = 6k+9 = 3*(2k+3). That is, last of the three consecutive odd numbers is divisible by 3 and so it is not a prime. Thus for any three consecutive odd numbers greater than 3, one of them is divisible by 3 and therefore the three cannot all be prime.
They are 3k/6k and 2k/6k where k is any non-zero integer.
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)
2k + 3 = 11(- 3 from both sides)2k = 8(divide both sides by 2)k = 4
2k = 5k-30 Subtract 5k from both sides: -3k = -30 Divide both sides by -3 to find the value of k remembering that a minus divided into a minus becomes a plus: k = 10