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3 years ago

What is the product in lowest terms?

2 answers:

6 0

#1. D. $\frac{9}{14}* \frac{-7}{12} = \frac{-63}{168}= \frac{-3}{8}$

#2. A.  $\frac{1}{3} of \frac{3}{4}=0.25= \frac{1}{4}$

brilliants [131]3 years ago

3 0

C
because you would take 3/4 and multiply it by 3/3 to get the denominator to 12
and you would multiply 1/3 by 4/4 again to get the denominator to 12.
You are getting the denominator to 12 because 12 is 3 and 4s Most Common multiple!
Now that you have multiplied them you should have 9/12 and 4/12
Then you subtract 4 from 9 to get 5.
So the answer is 5/12!

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Find three fractions whose product is -5/24.

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Answer:

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Step-by-step explanation:

-5/8 * 1/3 = -5/24

-5/12 * 1/2 = -5/24

-5/6 * 1/4 = -5/24.

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3 years ago

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Step-by-step explanation:

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3 years ago

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Find the square: (5x - 4y)2 (and please put the process of solving the equation)

satela [25.4K]

$(5x-4y)^2$
$(5x-4y)(5x-4y)$

Multiply each term in one bracket in turn by the terms in the other bracket

$(5x)(5x) = 25x^2$
$(5x)(-4y)=-20xy$
$(-4y)(5x)=-20xy$
$(-4y)(-4y)=16y^2$

Put them together we have
$25x^2-20xy-20xy-16y^2$
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3 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

Select the outlier in the data set.

Katarina [22]

Answer:

Step-by-step explanation:

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6 0

2 years ago