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Gnesinka [82]
3 years ago
12

Find the diameter and circumference 6.2 mi PLS HELP I NEED AN ANSWER PLS

Mathematics
1 answer:
Colt1911 [192]3 years ago
8 0

Answer:

diameter: 2.4 mi

circumference = 38.96 mi

Step-by-step explanation:

d=2r=2·6.2=12.4

C=2πr=2·π·6.2≈38.95575

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Answer:

Step-by-step explanation:

5x+10-7=-32

5x+3=-32

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5x=-35

X=-35/5

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5(-7+2)-7=32

5(-5)-7=32

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Consider all 5 letter "words" made from the full English alphabet. (a) How many of these words are there total? (b) How many of
VARVARA [1.3K]

Answer:

a) There are 11,881,336 of these words in total.

b) There are 7,893,600 of these words with no repeated letters.

c) 896,376 of these words start with an a or end with a z or both

Step-by-step explanation:

Our words have the following format:

L1 - L2 - L3 - L4 - L5

In which L1 is the first letter, L2 the second letter, etc...

There are 26 letters in the English alphabet.

(a) How many of these words are there total?

Each of L1, L2, L3, L4 and L5 have 26 possible options.

So there are 26^{5} = 11,881,336 of these words total

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At the second letter, the first one cannot be repeated, so L2 = L1 - 1 = 25.

At the third letter, nor the first nor the second one can be repeated, so L3 = L1 - 2 = 24

This logic applies until L5

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(c) How many of these words start with an a or end with a z or both (repeated letters are allowed)?

T = T_{1} + T_{2} + T_{3}

T_{1} is the number of words that start with an a and do not end with z. So we have

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The first letter can only be a, and the last one cannot be z. So:

T_{1} = 26^{3}*25 = 439,400

T_{2} is the number of words that start with any letter other than a and end with z. So we have

25 - 26 - 26 - 25 - 1

The first letter can be any of them, other than a, and the last can only be z. So:

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