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tia_tia [17]
3 years ago
12

The total number of seconds in a year is 32,000,000 when rounded to the nearest million. It is 31,500,000 when rounded to the ne

arest hundred thousand. Which could be the total number of seconds in a year? A. 31,092,000 seconds B. 31,476,000 seconds C. 31,536,000 seconds D. 31,592,000 seconds
Mathematics
1 answer:
babymother [125]3 years ago
3 0
Your answer will be B.
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Negative 5 is fewer than 6 less than a number k
lara [203]

Answer:

k > 1 is the solution.

Step-by-step explanation:

We are given that 'Negative 5 is fewer than 6 less than a number k', so algebraically it means that - 5 < k - 6.

Then this is a linear inequality of single variable k and we have to solve it.

Now, we have

- 5 < k - 6

⇒ - 5 + 6 < k - 6 + 6 {Adding 6 to both sides}

⇒ 1 < k

⇒ k > 1

This is the solution of the given inequality. (Answer)

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3 years ago
Given: KLMN is a trapezoid, m∠K = 90°
valentina_108 [34]
A=162sq......... bless up
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3 years ago
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CHANCE TO WIN BRAINLIEST! How much time does the average person spend on their phone (hours)?
soldi70 [24.7K]
8 hours and 41 mins per day!
(About 9 hours)
Hoped I helped!

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3 years ago
Find the value of x. *<br> Please help I’m really confused
mezya [45]

Due to the Triangle Angle Sum Theorem, we know that the sum of the interior angles of a triangle is 180 degrees.

90 + 2x - 2 + x + 5 = 180 degrees.

Combine like terms.

3x + 93 = 180

Subtract 93 from both sides.

3x = 87

Divide both sides by 3

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4 0
3 years ago
A certain organization recommends the use of passwords with the following​ format: vowel commavowel, consonant comma consonant c
zimovet [89]

The password format is = inpray93

vowel , consonant, consonant, consonant, vowel, consonant, number, number.

There are 21 consonants, 5 vowels and 10 number choices (0-9)

Since, there are 21 consonants and repetition is allowed so for 2nd, 3rd, 4th, and 6th positions 21 choices are available. For 1st and 5th positions, 5 choices  are available and for 7th and 8th position there are 10 choices (0 to 9).

Passwords are not case sensitive which means upper case and lower case letters are same.

So, the number of passwords can be =

5\times21\times21\times21\times5\times21\times10\times10

= 486202500

Part (b) question is not given. But it can be opposite to the first part i.e when the password letters are case sensitive.

This means all the letters are different.

So, number of passwords will be =

10\times42\times42\times42\times10\times42\times10\times10

= 31116960000

4 0
3 years ago
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