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This math problem equals out to be -22. This is partly learned in ninth grade math up to 12th grade. This math is used out of school too.

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79 + k = 84 84 - 79 = k k = 5

7(k + 5)(k - 4)

k = 2 7 + (5 x 2) = 17 (8 x 2) + 1 = 17

3k-5=7k+73k-7k=7+5-4k=12k=-3

2k + 5

It really does not. But is you really must, it is -13k/k where k is any non zero integer.

26-7k=6k First, add 7k to get k on one side. 26-7k+7k=6k+7k 26=13k Divide both sides by 13 to get k by itself. 26/13=13k/13 2=k The answer is 2.

It simplifies to 7k = 49 which gives k = 7.

So K > 93...

You don't need the q.f. That factors to (k + 6)(k - 1), so k = 1, -6 but if you insist... Plugging in the values for a, b and c leads us to (-5 plus or minus 7 times the square root of 1) divided by 2 (-5 + 7) ÷ 2 = 1 (-5 - 7) ÷ 2 = -6 but you knew that already. No the problem asked me to do it using the quadratic formula ...any way thanks

Equation: x^2 +2kx +10x +k^2 +5 = 0 Using the discriminant: (2k +10)^2 -4*1*(k^2 +5) = 0 Solving the discriminant: k = -2

K+15

7

k=4

==> 3k+14=k ==>3k-k=-14 ==>2k=-14 ==>k=-7 ans...

K-Ci is about 5' 7"

k = 5

16 - 3k + 5 and k is 4, then 16 -12 + 5 = 9

k + 6k - 13 = 177k = 30k = 30/7k = 4.2857

Using the discriminant of b^2 -4ac = 0 the value of k works out as -2

35 + 2k = 9k subtract 2k from each side 35 + 2k - 2k = 9k - 2k 35 = 7k divide both sides integers by 7 35/7 = (7/7)k 5 = k check in original equation 35 + 2(5) = 9(5) 35 + 10 = 45 45 = 45 checks out

When factored it is: (3k+5)(k-1)

35.

5

First, factor out the k. So, k(4+2) = 30 or 6k = 30 Then divide by 6. k = 30/6 = 5 So k = 5.