(2k + 3) (2k + 3)
2k - 3l
-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
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.
Take 3k from both sides to get k-3 equals 4, so k is 7
If by K2 you are not referring to the mountain, the distributive property of multiplication tells us K * 2 = 2 * K.Therefore: K2 + 2K + 4K = 2K + 2K + 4K = 8K
(2k + 3) (2k + 3)
It can be simplified to: 9k
2k - 3l
-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 + 24 = 6k - 10 subtract 4k from each side 4k - 4k + 24 = 6k - 4k - 10 24 = 2k - 10 add 10 to each side 10 + 24 = 2k - 10 + 10 34 = 2k divide each side integers by 2 17 = k ------------check 4(17) + 24 = 6(17) - 10 68 + 24 = 102 - 10 92 = 92 checks
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.
q=4k+4uImproved Answer:-If: Q = 4k+4uThen: k = (4u-Q)/-4
1k/2k, 2k/4k, 3k/6k and so on.
4k-91h
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
what is -5+3+4k=-115