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3 years ago

The explicit rule for an arithmetic sequence is an=20/3+1/3(n-1). What is the value of the 89th term?

1 answer:

5 0

$\bf a_n=\cfrac{20}{3}+\cfrac{1}{3}(n-1) \\\\[-0.35em] ~\dotfill\\\\ \stackrel{89^{th}~term~\hfill }{a_{89}=\cfrac{20}{3}+\cfrac{1}{3}(89-1)}\implies a_{89}=\cfrac{20}{3}+\cfrac{1}{3}(88)\implies a_{89}=\cfrac{20}{3}+\cfrac{88}{3} \\\\\\ a_{89}=\cfrac{20+88}{3}\implies a_{89}=\cfrac{108}{3}\implies a_{89}=36$

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Solve for p. <br><br> i = prt<br><br> p = i-rt<br> p = i/(rt)<br> p = (rt)/i

Debora [2.8K]

Answer:

Step-by-step explanation:

$i=prt\to prt=i\qquad\text{divide both sides by}\ rt\neq0\\\\\dfrac{prt}{rt}=\dfrac{i}{rt}\\\\p=\dfrac{i}{rt}$

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3 years ago

The system of equations 2y = 14 - 2x and y=-x+ 7 is

Ratling [72]

Answer:

Infinitely many solutions

Step-by-step explanation:

2y=14-2x

y=-x+7

------------

2(-x+7)=14-2x

-2x+14=14-2x

-2x-(-2x)=14-14

-2x+2x=0

0=0

infinitely many solutions

6 0

4 years ago

Read 2 more answers

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

3 years ago

lan ran 24 laps in a charity run and then walked 0.2 kilometer to the presentation table the total; distance lan traveled was 29

QveST [7]

Answer:

Step-by-step explanation:

1.225km is equal to each laps

Step-by-step explanation:

Since Kristy total distance traveled = 29.6km

And Kristy ran 24 laps and later walk 0.2km. it simply implies that

Total distance traveled = distance covered running + distance covered walking

Since we know that

Total distance traveled = 29.6km

distance covered running = 24 laps

distance covered walking = 0.2km

Distance covered running in km = total distance traveled - distance covered walking

Distance covered running in km = 29.6km - 0.2km = 29.4km.

To now find the distance for each lap.

Since we have:

Distance covered running in km = 29.4km.

Distance covered running in lap = 24 laps

i.e

24 laps = 29.4km

1 lap = x

Use cross-multiple

24 laps * x = 29.4km × 1 lap

x = 29.4km / 24

x = 1.225km

Therefore

1.225km is equal to each laps.

8 0

3 years ago

What two numbers add up two 40 but multiply to get 100

DerKrebs [107]

The two numbers are illustrations of system of equations

The two numbers are 2.7 and 37.3

<h3>How to determine the two numbers?</h3>

Let the numbers be x and y.

So, we have the following equations

x + y = 40

xy = 100

To determine the value of the variables, we simply plot the graph of both equations

From the graph, we have the following result

(x,y) = (2.7, 37.3)

Hence, the two numbers are 2.7 and 37.3

Read more about system of equations at:

brainly.com/question/14323743

3 0

2 years ago