Lenmana started studying how the number of branches on her tree grows over time. The number of branches increases by a factor of 11/6 every 3.5 years, and can be modeled by a function, N, which depends on the amount of time, t (in years).

When Lenmana began the study, her tree had 48 branches.

Write a function that models the number of branches t years since Lenmana began studying her tree.

N(t)=

Step-by-step explanation:

The formula for exponential growth is y = ab^x.

To write this equation, we know it has to start with 48 (which is the variable a). We need to add the rate of growth. This is 11/6 (which is variable b). But we also need to account for the "every 3.5 years" part, so divide the x as an exponent by 3.5.

N(t) = 48 * 11/6^(t/3.5)

This equation is easy to test, and it's a good idea to test it after you write it. For example, after 3.5 years we know that it should have 48*11/6 branches. Does our equation work? Yes.

48*(11/6)^t/3.5

Step-by-step explanation:

So lets put together this function like a nice PBJ. First they say that the tree had 48 branches. Knowing this, we can start the equatin of like this:

F(x) = 48 *

This is because we know 48 is the number being increased over time, and this number will have to be multiplied, since adding/subtracting/division is something I don't rly want to do.

So, we also know that the 48 branches increase by 11/6. Now we can add this to the function:

F(x) = 48 * (11/6)

This is because 11/6 is what will always be added to 48. The reason I put this in parentathese is so it a cleaner equation and cannot be misunderstood by a calculator or person.

So in addition, we are told how long it takes to increase the 48 by 11/6. This is 3.5 years. Remember tho, that the problem askes for the answer in a single year. So lets find what the increase is in 1 year. T equals years:

11/6 ^t(1 year)/3.5 years = the increase of 1 year.

Cleaning this up, it looks like(note that the parethansese are not needed below):

(11/6)^t/3.5

Now, lets just plug this lil number into the end of our function and we should see(remember that (11/6) was already included in the function so this does not need to be plugged into the end of our function):

48 * (11/6)^t/3.5

So this is your end function.

Intrested to see how it works? Lets plug in a number. I like the number 7, since it will plug into the 3.5 nicly and give us a clean exponent:

48*(11/6)^7/3.5

=

48*(11/6)^2

=(plug into calculator)

161.34

That looks correct.

Ti⊂k∫∈s ω∅∅p :)