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Q: What is 3 plus the product of 14 and k?

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

If: k+3=-3 Then: k=-3-3 k=-6

If: 12k = 84 then k = 7

(k*m)3 = k3*m3

One triplet (k) and two pairs (h and number). 3j has no like term.

k = 14/43

k + 5k = 6k 2 x 3 x k

K3PO4 + 3HCl -> 3KCl + H3PO4 Since K on the reactant side has 3 potassium atoms, K on the product side should also have 3 potassium atoms to balance the equation. Since you put the 3 on KCl of the product side, another 3 has to go on the Cl on the reactant side which also matches the 3 hydrogen atoms on the product side in H3PO4. If you check, the equation is now balanced. Everything that appears on the left, equally appears on the right

K subscrpit 3PO subscript 4= K3PO4

3k-1=k+2 2k=3 k=3/2=1.5

k = -3; factors are (9x + 3)(x - 1)

m+6n

j=-3 and k=2

2k + 5

K/3 + k/4 = 1 LCD=12 *divide lcd by denominator* K(4) + K(3) = 12(1) 4k + 3k = 12 7k = 12 k=12/7

60k plus k is 61k

K+

y = 3x + 2x^2 + k where k is any number

Yes, they are exactly the same, both of them increment k in 1.

Print "Type the upper limit (n) ?" Input n K = -1 WHILE K < = n K = K + 2 Sum = Sum + K WEND Print "The sum of all odd numbers up to "; n; "is "; Sum

Take 3k from both sides to get k-3 equals 4, so k is 7

Taking all terms and conditions into consideration a quadratic equation can be finally formed as such that 3k^2 +8k +4 = 0 whose solutions are k = -2/3 or k = -2 Check: 3(-2/3)^2 +8(-2/3) +4 = 0 Check: 3(-2)^2 +8(-2) +4 = 0

2^3 x 3 x 11

To factor this, find two numbers whose product is 81 and sum 18. These two numbers are therefore 9 and 9. So we can rewrite the polynomial in its factored form as follows: (k+9)(k+9), or "k plus 9 squared"