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Ber [7]
3 years ago
15

The length of a rectangle is 18 centimeters less than five times its width. its area is 35 square centimeters. find the dimensio

ns of the rectangle.
Mathematics
1 answer:
Studentka2010 [4]3 years ago
5 0
35 = 5*7
7 = 5*5 -18

The dimensions of the rectangle are: length = 7 cm, width = 5 cm.

_____
It is handy to know your times tables.

___
If you want, you can write a couple of equations and solve them.
.. L = 5W -18
.. LW = 35
Substituting for L, we have
.. (5W -18)W = 35
.. 5W^2 -18W -35 = 0 . . . . put in standard form
.. (W -5)*(5W +7) = 0 . . . . . factor
.. W = 5 or -7/5 . . . . . . . . . we are only interested in the positive solution

The width of the rectangle is 5 cm; its length is 7 cm.
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2x+y= -2, x + 2y = 2

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Is 5 divided by 4 a rational number?
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Determine how many different computer passwords are possible if (a) digits and letter so can be repeated and (b) digits and lett
Lerok [7]

Answer:

a) 1,188,137,600 different passwords.

b) 710,424,000 different passwords.

Step-by-step explanation:

We have a code of 2 digits followed by 5 letters.

First, the total number of digits is 10

The total number of letters is 26.

Then:

a) Digits and letters can be repeated.

Here we need to count the number of options for each selection.

For the first digit, we have 10 options.

For the second digit, we have 10 options.

For the first letter, we have 26 options.

For the second letter, we have 26 options.

For the third letter, we have 26 options.

For the fourth letter, we have 26 options.

For the fifth letter, we have 26 options.

The total number of combinations will be equal to the product of the number of options. We get:

Combinations = 10*10*26*26*26*26*26 = (10^2)*(26^5) = 1,188,137,600

This means that we have  1,188,137,600 different possible passwords.

b) Digits and letters can not be repeated.

We start in the same way as above:

For the first digit, we have 10 options.

For the second digit, we have 9 options, because one is already used.

For the first letter, we have 26 options.

For the second letter, we have 25 options, because one letter is already used.

For the third letter, we have 24 options, because 2 letters are already used.

For the fourth letter, we have 23 options, because 3 letters are already used.

For the fifth letter, we have 22 options, because 4 letters are already used.

Then the total number of combinations is:

Combinations = 10*9*26*25*24*23*22 = 710,424,000

So if we can not repeat digits nor letters, we can make 710,424,000 different passwords,

8 0
3 years ago
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