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mina [271]
3 years ago
13

Write the equation of a line that has a slope 1/3 and y-intercept 4.

Mathematics
1 answer:
tangare [24]3 years ago
5 0
Y=mx+b, where m represents the slope and b represents the y-intercept. If you plug in the numbers, you get the equation, y=(1/3)x+4
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Help PLEASE I don't get this giving many points
Tom [10]

Answer:

x=20

Step-by-step explanation:

7 0
3 years ago
What is the answer to this​
Alborosie
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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
A water tank already contained 24 gallons of water before a pump was switched to fill it. The pump started at 10:00 am and by 10
grigory [225]

Answer:

The equation to represent  the amount of water that the tank contains (W) after the pump was working for T hours can be given as:

W=32T+24

Step-by-step explanation:

Given:

Water tank initially contained = 24 gallons of water

The pump to fill the tank was switched to fill it.

From 10:00 AM to 10:30 AM the tank contained 40 gallons of water.

By 1:00 PM the tank contained 120 gallons of water.

To write an equation  which will tell the amount of water that the tank contains (W) after the pump was working for T hours

Solution:

Initially, gallons of water in tank = 24

From 10:00 AM to 10:30 AM the tank had 40 gallons of water.

Thus, in half an hour the pump filled = 40-24 = 16 gallons

Using unitary method:

If in \frac{1}{2} of an hour water filled = 16 gallons

So, in 1 hour the water filled will be = 16\times 2 = 32 gallons

In 3 hours the water filled will be = 32\times 3 = 96 gallons

Thus, by 1:00 PM the water in the tank will be = 96+24 = 120 gallons

This confirms the data given for 1:00 PM.

This shows that the rate at  which the pump fills the tank is 32 gallons of water per hour.

The equation to represent  the amount of water that the tank contains (W) after the pump was working for T hours can be given as:

W=32T+24

4 0
3 years ago
I don’t know if this is correct , please help !!!!!! Will mark brainliest !!!!!!!!!!!!
Arisa [49]

Answer:

change the negative to not domain and the zero to in domain

Step-by-step explanation

7 0
2 years ago
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