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3 years ago

The absolute value of a number is

Mathematics

2 answers:

DENIUS [597]3 years ago

5 0

Answer:

The absolute value of a number is always positive. So that means you circled the correct one. GOOD JOB!!!!!!!!!!!!!!!!!!!!!!!

Step-by-step explanation:

Elanso [62]3 years ago

5 0

absolute value will always be positive

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Question 5 (1 point)

Bess [88]

Answer:

x > -3

Step-by-step explanation:

-3x + 6 < 15

-3x < 15 - 6

-3x < 9

-x < 3

x > -3

3 0

2 years ago

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What is the total floor area in inches of a rectangular floor 8ft. by 6ft.?

Anestetic [448]

Answer:the answeer is

= 72 ft³

Step-by-step explanation:

Multiply the width of the wall by its height. So one of the walls is 80 square feet (10 feet wide x 8 feet high) and the other is 96 square feet (12 feet x 8 feet). If you need the total square footage of the walls - for figuring paint or wallpaper for example - you can simplify the calculation by first adding all the wall lengths together, then multiplying by the height (10 + 12 + 10 + 12 = 44 x 8 = 352 square feet of total wall area).

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2 years ago

How to solve word problems

Lapatulllka [165]

<span>Read the problem entirely.
Get a feel for the whole problem.
List information and the variables you identify.
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6 0

3 years ago

Read 2 more answers

Suppose you always eat a particular brand of cookies out of the standard sized package it comes in from the store. You notice th

Blababa [14]

Answer:

There are 91 cookies in the package of cookies

Step-by-step explanation:

The given parameter for the number of cookies are;

The number of cookies left when eating 3 at a time = 1 cookie

The number of cookies left when eating 5 at a time = 1 cookie

The number of cookies left when eating 7 cookies at a time = No cookie leftover

Let 'n' represent the number of cookies in the pack, let 'a', 'b', and 'c' be the multiples of 3s, 5s, and 7s in 'n', respectively, we have;

The number of cookies in the pack, n < 100

Therefore, we have;

3·a = n - 1

5·b = n - 1

7·c = n

∴ n - 1 is a multiples of 3 and 5 less than 100 which are;

15, 30, 45, 60, 75, and 90

Therefore, the possible values of 'n' are'

n = n - 1 + 1 = 15 + 1 = 16, 30 + 1 = 31, 45 + 1 = 46, 60 + 1 = 61, 75 + 1 = 75, 90 + 1 = 91

Multiples of 7 includes;

7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, and 98

Therefore, n = 91 which is the number that appear simultaneously in both search

The number of cookies in a package = 91 cookies

3 0

2 years ago

Lim n→∞[(n + n² + n³ + .... nⁿ)/(1ⁿ + 2ⁿ + 3ⁿ +....nⁿ)]

Schach [20]

Step-by-step explanation:

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\:\displaystyle\lim_{n \to \infty} \frac{n + {n}^{2} + {n}^{3} + - - + {n}^{n} }{ {1}^{n} + {2}^{n} + {3}^{n} + - - + {n}^{n} }$

To, evaluate this limit, let we simplify numerator and denominator individually.

So, Consider Numerator

$\rm :\longmapsto\:n + {n}^{2} + {n}^{3} + - - - + {n}^{n}$

Clearly, if forms a Geometric progression with first term n and common ratio n respectively.

So, using Sum of n terms of GP, we get

$\rm \: = \: \dfrac{n( {n}^{n} - 1)}{n - 1}$

$\rm \: = \: \dfrac{ {n}^{n} - 1}{1 - \dfrac{1}{n} }$

Now, Consider Denominator, we have

$\rm :\longmapsto\: {1}^{n} + {2}^{n} + {3}^{n} + - - - + {n}^{n}$

can be rewritten as

$\rm :\longmapsto\: {1}^{n} + {2}^{n} + {3}^{n} + - - - + {(n - 1)}^{n} + {n}^{n}$

$\rm \: = \: {n}^{n}\bigg[1 +\bigg[{\dfrac{n - 1}{n}\bigg]}^{n} + \bigg[{\dfrac{n - 2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]$

$\rm \: = \: {n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]$

Now, Consider

$\rm :\longmapsto\:\displaystyle\lim_{n \to \infty} \frac{n + {n}^{2} + {n}^{3} + - - + {n}^{n} }{ {1}^{n} + {2}^{n} + {3}^{n} + - - + {n}^{n} }$

So, on substituting the values evaluated above, we get

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{\dfrac{ {n}^{n} - 1}{1 - \dfrac{1}{n} }}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{ {n}^{n} - 1}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{ {n}^{n}\bigg[1 - \dfrac{1}{ {n}^{n} } \bigg]}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{\bigg[1 - \dfrac{1}{ {n}^{n} } \bigg]}{\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{1}{\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

Now, we know that,

$\red{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to \infty} \bigg[1 + \dfrac{k}{x} \bigg]^{x} = {e}^{k}}}}$

So, using this, we get

$\rm \: = \: \dfrac{1}{1 + {e}^{ - 1} + {e}^{ - 2} + - - - - \infty }$

Now, in denominator, its an infinite GP series with common ratio 1/e ( < 1 ) and first term 1, so using sum to infinite GP series, we have

$\rm \: = \: \dfrac{1}{\dfrac{1}{1 - \dfrac{1}{e} } }$

$\rm \: = \: \dfrac{1}{\dfrac{1}{ \dfrac{e - 1}{e} } }$

$\rm \: = \: \dfrac{1}{\dfrac{e}{e - 1} }$

$\rm \: = \: \dfrac{e - 1}{e}$

$\rm \: = \: 1 - \dfrac{1}{e}$

Hence,

$\boxed{\tt{ \displaystyle\lim_{n \to \infty} \frac{n + {n}^{2} + {n}^{3} + - - + {n}^{n} }{ {1}^{n} + {2}^{n} + {3}^{n} + - - + {n}^{n} } = \frac{e - 1}{e} = 1 - \frac{1}{e}}}$

3 0

2 years ago

The absolute value of a number is​

The absolute value of a number is