ONE QUESTION
here, I think to conjecture a linear relationship between x= number of months, and y = number of sunscription, namely: y=ax+b
so we have to solve the subsequent system: \[\left\{ \begin{gathered} 5730 = 5a + b \hfill \\ 6022 = 7a + b \hfill \\ \end{gathered} \right.\]
where a and b have to be determined, using, for example the substitution method
hint: I solve the first equation, and I get: b= 5730-5a
next I substitute that value for b inthe secon equation, as below: \[6022 = 7a + 5730 - 5a\]
what is a?
one sec im doing it
146
ok!
now I substitutte that value for a into the first equation, for example, and I get: \[b = 5730 - 5a = 5730 - 5 \times 146 = ...?\]
then what is b?
5730-5a
yes, and a = 146, so: b = 5730 - 5* 146= 5730 - 730= ...?
5000?
perfect!
so our relationship, is: \[y = 146x + 5000\]
sorry i confused myself but i understand now
ok!
solve that ^
now, we have to set x= 10 into the previous formula, and then we have to compute y: \[y = 146x + 5000 = 146 \times 10 + 5000 = ...?\]
what is y?
hint: \[146 \times 10 + 5000 = 1460 + 5000 = ...?\]
im not sure dude
why?
do i just solve that straight out
where do i start?
in order to answer to your problem, we have to write the right function which models your exercise. Now, I think you are studying the linear relationships, so I have conjectured a linear relationship between the months x and the subscription y. Then your starting point is this function: y= ax+b
Now, using your data, we have to determine the values of a and b. We got: a = 146, and b = 5000
finally we have to write how many are the Subscriptions after 10 months. In order to that we have to set x=10 into our formula, namely: y = 146*10 +5000 what do you get is the right answer to your problem
yes but what i dont get is what do i solve for here ->b = 5730 - 5* 146= 5730 - 730=
there you are finding the value of the coefficient b: b= 5730-730=5000
so the function which models your problem is: y = 146x+5000
so for x or y?
i got 146
that's the value of a, furthermore, you have to evaluate y when x=10
dang this one is Hard
oh i se one sec
6460
SO D :)
congrats! you got the right answer
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