Square root of 48a^5 *b over square root of 12ab. Simplified

1 answer:

4 0

$\frac{\sqrt{48a^5b}}{\sqrt{12ab}}=\frac{\sqrt{16\times3}\cdot a^{\frac{5}{2}}\cdot b^{\frac{1}{2}}}{{\sqrt{4\times3}}\cdot a^{\frac{1}{2}}b^{\frac{1}{2}}}=\frac{4\sqrt{3}\cdot a^{\frac{5}{2}}}{{2\sqrt{3}}\cdot a^{\frac{1}{2}}}=2a^{\frac{5}{2}-\frac{1}{2}}=2a^2$

You might be interested in

A computer system uses passwords that are six characters, and each character is one of the 26 letters (a–z) or 10 integers (0–9)

Blababa [14]

First of all, since we have 36 characters available per spot (26 letters and 10 digits), and we have 6 spots, we have a total of

$36^6$

possible passwords.

Event A happens if the password starts with either a, e, i, o or u. If we fix the first character, we're left with 36 characters available for each of the remaining 5 spots, leading to a total of

$5\cdot 36^5$

possible passwords.

So, the probability of event A, computed as the ratio between "good" cases and all possible cases, is

$\dfrac{5\cdot 36^5}{36^6}=\dfrac{5}{36}$

Event B works exactly the same, since we're fixing the last spot, leaving 36 characters available for each of the first 5 spots. So, we have

$P(A)=P(B)=\dfrac{5}{36}$

As for the intersection, we want the first character to be a vowel, and the last character to be an even digits. There are 25 passwords satisfying this request:

$axxxx0,\ axxxx2,\ axxxx4,\ axxxx6,\ axxxx8$