7(k + 5)(k - 4)
Wiki User
∙ 8y ago-6K+7K=1K or K
-6k + 7k = k
k(7k + 9)
It simplifies to 7k = 49 which gives k = 7.
No because 0 is additive inverse the multiplicative inverse of a number is when you multiply that number by the m. inverse and still get that number.
35k2 - 22k + 3 = 0 35k2 - 15k - 7k + 3 = 0 5k(7k - 3) - (7k - 3) = 0 (5k - 1)(7k - 3) = 0 5k -1 = 0 or 7k - 3 = 0 5k = 1 or 7k = 3 k = 1/5 or k = 3/7
-6K+7K=1K or K
-6k + 7k = k
Put it into two binomilals that multipy together to create the polynomial. For example: 5K(squared)-2k-7 is factored out as: (5k+1)(-7K+1)
-6k + 7k = k or 7k- 6k = k or Factor out 'k' k(-6 + 7) = k(1) = k
1k
7k + 4 + 4k - 9 - 5k (7k + 4k - 5k) + (4 - 9) 6k + -5 6k - 5
3k-5=7k+73k-7k=7+5-4k=12k=-3
-6k+7k = 1k or normally k on its own.
The answer is 13k .
3k+4m+2
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