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3 years ago

In how many ways can a password be made using four letters from the 26 letter English alphabet letters can’t be repeated

Mathematics

1 answer:

valentinak56 [21]3 years ago

7 0

Answer:

If you go by normal method, no. of possible passwords that can be created with 26 alphabets are 26X26X26X26 = 456,976

Step-by-step explanation:

. If you do not want to repeat the alphabets, you can use combinations. i.e. C(26, 4) or 26C4 = 358,800 ways

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3 years ago

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4 years ago

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3 years ago

Send help please help :D

Svetllana [295]

Answer:

A) Slope =

$m = \frac{1}{4} x$

explanation:

Take 2 random points off the line, I used (4,1) & (8,2)

The formula to find the slope when given 2 points is

$m = \frac{rise}{run} = \frac{y2 - y1}{x2 - x1}$

so in this case x1 & y1 are assigned to (4,1)

4 = x1
1 = y1

and x2 & y2 are assigned to (8,2)

8 = x2
2 = y2

so substitute the numbers into the formula and solve !

$m= \frac{y2 - y1}{x2 - x1} = \frac{2 - 1}{8 - 4} = \frac{1}{4}$

this will result in the answer I got above!!

B) yes it does show a constant rate of change cause it's increasing in 1/4x intervals

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5 0

3 years ago

X^-2+4x^-1+3=0 solve by making appropriate substitution

Alex

ANSWER

$x = - 1 \: or \: x = - \frac{1}{ 3}$

EXPLANATION

The given equation is:

${x}^{ - 2} + 4 {x}^{ - 1} + 3 = 0$

Recall that:

${a}^{ - m} = \frac{1}{ {a}^{m} }$

$\frac{1}{ {x}^{2} } + \frac{4}{x} + 3 = 0$

${( \frac{1}{x} )}^{2} + 4( \frac{1}{x} ) + 3 = 0$

Let

$u = \frac{1}{x}$

Our equation then becomes:

${u}^{2} + 4u + 3 = 0$

The factors of 3 that add up to 4 are:

${u}^{2} + 3u + u + 3$

$u(u + 3) + 1(u + 3) = 0$

$(u + 1)(u + 3) = 0$

$u + 1 = 0 \: or \: u + 3 = 0$

$u = - 1 \: or \: u = - 3$

This implies that:

$\frac{1}{x} = - 1 \: or \: \frac{1}{x} = - 3$

$x = - 1 \: or \: x = - \frac{1}{ 3}$

7 0

3 years ago