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exponentials, please help!

A journalist observes that the total number of twitter users, between 2011 and 2014, can be modelled by the linear equation
u = 64.9t-660 (11<t<14)
where u is the total number of Twitter users in millions, and t is the number of years since the start of 2000.
1) what is the number of users at the start of 2012
2) calculate the year in which the total number of users reaches 150 million
3) write down the gradient of the straight line represented by the equation
Original post by sue2win
A journalist observes that the total number of twitter users, between 2011 and 2014, can be modelled by the linear equation
u = 64.9t-660 (11<t<14)
where u is the total number of Twitter users in millions, and t is the number of years since the start of 2000.
1) what is the number of users at the start of 2012
2) calculate the year in which the total number of users reaches 150 million
3) write down the gradient of the straight line represented by the equation


What have you tried?
Original post by sue2win
A journalist observes that the total number of twitter users, between 2011 and 2014, can be modelled by the linear equation
u = 64.9t-660 (11<t<14)
where u is the total number of Twitter users in millions, and t is the number of years since the start of 2000.
1) what is the number of users at the start of 2012
2) calculate the year in which the total number of users reaches 150 million
3) write down the gradient of the straight line represented by the equation


What have you tried so far? :h:
Reply 3
I cannot seem to get my head around it at all :frown: - any advice on how to work it out would be greatly appreciated.
Original post by sue2win
I cannot seem to get my head around it at all :frown: - any advice on how to work it out would be greatly appreciated.


Well it's a linear expression and t is the amount of years after 2012. So what would t be at 2012? Use that and sub it in for the first one. For the second one, equate the linear expression to 150 and solve for t. For third one, well it's pretty obvious.

There are no exponentials here.
Reply 5
Original post by RDKGames
Well it's a linear expression and t is the amount of years after 2012. So what would t be at 2012? Use that and sub it in for the first one. For the second one, equate the linear expression to 150 and solve for t. For third one, well it's pretty obvious.

There are no exponentials here.


so would I do 64.9^12 for the first one?
Original post by sue2win
so would I do 64.9^12 for the first one?


Errr... Okay clear this up. Did you mean the original equation to be u=64.9t660,11<t<14u=64.9^t-660 , 11<t<14 ???
Reply 7
Original post by RDKGames
Errr... Okay clear this up. Did you mean the original equation to be u=64.9t660,11<t<14u=64.9^t-660 , 11<t<14 ???


No, I have looked at the question and it is as I originally wrote it, so how do I work out what t is at 2012? because the question gives the impression that the answer should be in millions???
Original post by sue2win
No, I have looked at the question and it is as I originally wrote it, so how do I work out what t is at 2012? because the question gives the impression that the answer should be in millions???


Then just sub t=12, like you've done, into the given equation. The answer would like 3 digits followed by any decimals, which are in the millions as stated by the information.
Reply 9
Original post by RDKGames
Then just sub t=12, like you've done, into the given equation. The answer would like 3 digits followed by any decimals, which are in the millions as stated by the information.




ill have ago, thank you
Reply 10
Original post by sue2win
No, I have looked at the question and it is as I originally wrote it, so how do I work out what t is at 2012? because the question gives the impression that the answer should be in millions???


When it says that uu is in millions it means those are the units of u.

So if you put in t=11t=11 and get u=53.9u = 53.9 this means that there are 53.9 million twitter users. Get it?
Reply 11
Original post by Zacken
When it says that uu is in millions it means those are the units of u.

So if you put in t=11t=11 and get u=53.9u = 53.9 this means that there are 53.9 million twitter users. Get it?


so, u = 64.9 x 11 - 660

thank you !!
Reply 12
Original post by sue2win
so, u = 64.9 x 11 - 660

thank you !!


to calculate the year in which the total number reaches 150 million would I;

t = 150/660 x 64.9
t=14.75 ??
Original post by sue2win
to calculate the year in which the total number reaches 150 million would I;

t = 150/660 x 64.9
t=14.75 ??


Nope. Try again. u=150u=150 for that.
Reply 14
Original post by RDKGames
Nope. Try again. u=150u=150 for that.


hi all

Stuck on same question as sue2 win must be doing same maths course. I know the answer which is 2nd half of 2012 or 12.5 but i cant derive the answer. Any help on getting to the solution please.

Thanks
Eggie
Original post by eggie
hi all

Stuck on same question as sue2 win must be doing same maths course. I know the answer which is 2nd half of 2012 or 12.5 but i cant derive the answer. Any help on getting to the solution please.

Thanks
Eggie


150=64.9t660150=64.9t-660 and solve for t.
Original post by sue2win
A journalist observes that the total number of twitter users, between 2011 and 2014, can be modelled by the linear equation
u = 64.9t-660 (11<t<14)
where u is the total number of Twitter users in millions, and t is the number of years since the start of 2000.
1) what is the number of users at the start of 2012
2) calculate the year in which the total number of users reaches 150 million
3) write down the gradient of the straight line represented by the equation


Here's your equation formatted neatly:

[br]u=64.9t660[br] [br]u = 64.9t\,\displaystyle{-}\,660[br]

where u is the total number of Twitter users (in millions)
where t is the number of years, since the start of 2000
where 11 < t < 14 (i.e. The equation only works for years between 2011 and 2014)


Question 1
What is the number of users at the start of 2012?

2012 is 12 years after 2000, therefore t = 12
Substitute this into the equation:

[br]u=64.9×12660[br][br]u = 64.9 \times\, 12 - 660[br]

Evaluate to find u.


Question 2
Calculate the year in which the total number of users reaches 150 million.

We know the number of users is 150 million, so u = 150, now we just find t:

[br]150=64.9t660[br][br]150 = 64.9t\,\displaystyle{-}\, 660[br]

Rearrange to find t.
Hint about rounding your answer for t: The question asks for "the year in which" Twitter has 150 million users, so it doesn't matter whereabouts in that year it happened.


Question 3
Write down the gradient of the straight line represented by the equation.

The equation is called u, and it is in terms of t. Therefore the gradient is:

Unparseable latex formula:

[br]gradient = \displaystyle{\frac{du}{dt}}\\[br]



Another way to find the gradient, is knowing that u is linear, meaning it is a straight line, and it's in the form y=mx+cy = mx + c You should know fr GCSE that m is the gradient of a straight line.
(edited 7 years ago)
Reply 17
Hi all

A journalist observes that the total number of Twitter users, between 2011 and
2014, can be modelled by the linear equation u = 64.9t−660 (11 t 14),
where u is the total number of Twitter users in millions, and t is the number of years since the start of 2000.

Can anyone tell me this please

Explain why the term 660 does not imply that there were ever a negative number of Twitter users at the start of 2000

Thanks
Eggie
Original post by eggie
Hi all

A journalist observes that the total number of Twitter users, between 2011 and
2014, can be modelled by the linear equation u = 64.9t−660 (11 t 14),
where u is the total number of Twitter users in millions, and t is the number of years since the start of 2000.

Can anyone tell me this please

Explain why the term 660 does not imply that there were ever a negative number of Twitter users at the start of 2000

Thanks
Eggie


The question tells you that 11 t 14 which means that the equation will ONLY work with values of t that are inside that range.

In 2000, t = 0, which is not in the range and so (as you have found) the equation does not work.

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