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vivado [14]
3 years ago
7

The identification code on a bank card consists of 1 digit followed by 2 letters. The code must meet the following conditions: T

he digit must be odd. The letters A, E, I, O, and U cannot be used. Letters cannot be used more than once.
Mathematics
1 answer:
jasenka [17]3 years ago
8 0

Answer:

For the code we have 3 selections.

The first selection is a digit that must be odd, so the options are {1, 3, 5, 7 ,9}

So we have 5 options.

The second selection is a letter from the set of all the letters (27) minus the set of the vowels (5)

So here we have 27 - 5 = 22 options

The third selection is also a letter from the previous set, but because each letter can be used only one time, and in the previous selection we already selected one of the letters, in this selection we have a letter less than in the previous selection.

Here we have 22 - 1 = 21 options.

The total number of combinations (of possible codes) is equal to the product of the number of options for each selection:

C = 5*22*21 = 2310.

There are 2310 different possible codes

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I need a written answer for all three questions please.
Setler79 [48]

The values of the trigonometry ratios are:

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<h3>How to solve the trigonometry ratios?</h3>

<u>1: sin α = -12/13 and tan α > 0, find cos α and cot α</u>

Because tan α > 0, then it means that cos α and sin α are negative

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sin²α + cos²α = 1

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Evaluate the difference

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<u>2: tan α = -12/5 for α in quadrant IV, find sec α and cot α</u>

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cot α = 1/tan α

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sec²α = 1 + (-12/5)²

Evaluate the squares

sec²α = 1 + 144/25

Evaluate the sum

sec²α = 169/25

Take the square root of both sides

sec α = 13/5

Hence, cot α = -5/12 and sec α = 13/5

Read more about trigonometry ratios at:

brainly.com/question/11967894

#SPJ1

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