There are 6 other positions in the string, each with 26 choices. So if you fix BO as the first two letters, there are $26^6$ possible strings that you can make.

If BO is at the end of the string, you still have $26^6$ possible strings.

Together, then, you have $2\times26^6$ possible strings.

5 0

3 years ago

What is the solution set of the equation 30/x^2-9 +1=5/x-3

gizmo_the_mogwai [7]

$\frac{30}{x^{2} - 9 + 1} = \frac{5}{x - 3} \\\frac{30}{x^{2} - 8} = \frac{5}{x - 3} \\5(x^{2} - 8) = 30(x - 3) \\5(x^{2}) - 5(8) = 30(x) - 30(3) \\5x^{2} - 40 = 30x - 90 \\5x^{2} + 50 = 30x \\5x^{2} - 30x + 50 = 0 \\5(x^{2}) - 5(6x) + 5(10) = 0 \\5(x^{2} - 6x + 10) = 0 \\\frac{5(x^{2} - 6x + 10)}{5} = \frac{0}{5} \\x^{2} - 6x + 10 = 0 \\x = \frac{-(-6) \± \sqrt{(-6)^{2} - 4(1)(10)}}{2(1)} \\x = \frac{6 \± \sqrt{36 - 40}}{2} \\x = \frac{6 \± \sqrt{-4}}{2} \\x = \frac{6 \± 2i}{2} \\x = 3 \± i \\x = 3 + i\ or\ x = 3 - i$

The answer is D.

8 0

3 years ago

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