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3 years ago

3(2b-4a+3c) if a =1/2,b=3, and c=-2/3

2 answers:

8 0

This is just substitution. so 3(2(3)-4(1/2)+3(-2/3)= 3(6-2-2)= 3(2) = 6. Basically you plug in the values they gave you for the variables and then just solve one step at a time

STatiana [176]3 years ago

4 0

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Subract (21x) - (-16 + 7x) can someone help me

koban [17]

The solution to (21x) - (-16 + 7x) is 14x + 16

<h3><u>Solution:</u></h3>

Given that (21x) - (-16 + 7x)

We have to evaluate the given expression

Let's look at the given expression

(21x) - (-16 + 7x)

Multiply the terms within second bracket with minus sign

<em><u>There are two simple rules to remember:</u></em>

When you multiply a negative number by a positive number then the product is always negative.

When you multiply two negative numbers or two positive numbers then the product is always positive

⇒ (21x) + 16 - 7x

Removing the parenthesis

⇒ 21x + 16 - 7x

Combining the like terms,

⇒ 14x + 16

Thus the solution is 14x + 16

7 0

2 years ago

Organize the data into frequency table

irga5000 [103]

Wheres the data? we cant answer without the data.

8 0

3 years ago

Twelve members of a band purchase one beverage each during a break at practice. Exactly 5 of the 12 members bought bottled water

Hoochie [10]

Answer:

Step-by-step explanation:

You have to use ratios. If 5 of 12 members bought water, then x of 168 members will buy water.

$\frac{5}{12} = \frac{x}{168}\\12x = 840\\x = 70$

70 members will buy bottled water.

8 0

3 years ago

Dave has $8000 to invest for 15 years. He finds a bank that offers an interest rate of 3.1% compounded monthly. How much money w

ioda

Dave will have $12,728 after 15 years, if he has $8000 to invest for 15 years. He finds a bank that offers an interest rate of 3.1% compounded monthly.

Step-by-step explanation:

The given is,

Investment = $ 8000

No. of years = 15 years

Interest rate, i = 3.1 %

( compounded monthly )

Step:1

For for calculating future value with compound interest monthly,

$A = P (1 +\frac{r}{n})^{nt}$ .................(1)

Where,

A = Future amount

P = Initial investment

r = Rate of interest

n = Number of compounding in a year

t = Time period

Step:2

From given values,

P = $8000

r = 3.1%

t = 15 years

n = 12 ( for monthly)

Equation (1) becomes,

$A = 8000( 1+\frac{0.031}{12} )^{(12)(15)}$

$= 8000 (1+0.002583)^{180}$

$= 8000(1.002583)^{180}$

$= 8000(1.591059)$

$=12728.48$

A = $ 12728.48

Result:

Dave will have $12,728 after 15 years, if he has $8000 to invest for 15 years. He finds a bank that offers an interest rate of 3.1% compounded monthly.

8 0

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