Remember in basic Algebra we learned that all variables can be substituted as 1's If needed. So lets do that:
-k - 4k = -1 - 4(1) = -5
So now we can rewrite this removing the 1's we put in:
-k - 4k = -5k
k=-2
The generic expression for this type of question is as follows: cx * x = cx2 Where c is a constant, and x is a variable. So 4k * k = 4k2.
-4k + 10 - k + 2 = 2k + 4k + 18Combine all the 'k' terms on the left side:-5k + 10 + 2 = 2k + 4k + 18Combine all the 'k' terms on the right side:-5k + 10 + 2 = 6k + 18Combine the numerical terms on the left side:-5k + 12 = 6k + 18Add 5k to each side:12 = 11k + 18Subtract 18 from each side:-6 = 11kDivide each side by 11:k = -6/11
4k = 28 Therefore, 28/4 = 7 k = 7
One thousand. ie. 4k is 4,000 140k is 140,000
4k-7 = 7 4k = 14 k = 3.5 or k = 7/2
31 = 3-4k 31-3 = -4k 28 = -4k Divide both sides of the equation by -4 to find the value of k: k = -7
8-4k = 40 (4k-8) - 8 = 40-8 4k = 32 4k/4 = 32/4 k =8
4k-64 = -60
q=4k+4uImproved Answer:-If: Q = 4k+4uThen: k = (4u-Q)/-4
-4k-2=10 -4k=10+2 -4k=12 k=-3
There are infinitely many of them.Take any positive integer k and let 0
x =√k+√k+√k…… ectx² = k + xx² - x = kx² - x + (1/2)² = k + (1/2)² factorise x² - x + (1/2)²(x + 1/2)² = k + 1/4(x + 1/2)² = (4k + 1)/4x + 1/2 = ± √(4k + 1)/2x = 1/2 ± √(4k + 1)/2it obviously has to be positivex = 1/2 + √(4k + 1)/2x = (1 + √(4k + 1))/2
J = K+6 K + J = 4K Replace J in the second equation K + (K + 6) = 4K K + K + 6 = 4K 2K + 6 = 4K Subtract 2K from each side 6 = 2K divide both sides by 2 3 = K So, Keisha is 3 and therefore John is 9
k = 2
k=-2
The generic expression for this type of question is as follows: cx * x = cx2 Where c is a constant, and x is a variable. So 4k * k = 4k2.