The posting reference column in the ledger is
2 answers:
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Answer:
76.
Step-by-step explanation:
It is given that N is a 2-digit number.
Last two digits of N^2 is the same as N.
We know that, a number is even if it ends with 0,2,4,6,8.
If 0 is in end then we get two zeros in the square of that number.
It is clear that, number should ends with 6 to get the same number at the end.
It is clear that last two digits of (76)^2 is the same as 76.
Therefore, the required number is 76.
Answer:
x = ± - 3
Explanation:
I'm assuming you want the solutions to that equation, so here goes! (If not, please comment.)
(x-3)(x+9)=27
Let's FOIL this all out and expand. (Remember: First, Outer, Inner, Last.)
x^2 + 9x - 3x - 27
(first+ inner + outer + last)
x^2 + 9x - 3x - 27 = 27
Combine like terms, and add 27 to both sides.
x^2 + 6x - 27 + 27 = 27 + 27
x^2 + 6x = 54
Let's complete the square, because factoring doesn't work, and because it's good practice.
x^2 + 6x + ___ = 54 + ____
In the blank we will put b/2 ^2 = 6/2 ^2 = 3^2 = 9 to complete the square.
x^2 + 6x + 9 = 54 + 9
Now we've got a perfect square factor:
(x + 3)^2 = 63
sqrt(x+3)^2 =
x + 3 = ±
x = ± - 3