If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains ***.kastatic.org** and ***.kasandbox.org** are unblocked.

Main content

Current time:0:00Total duration:4:21

CCSS.Math: , , , , , ,

oceans Sun fishes are well known for rapidly gaining a lot of weight on a diet based on jellyfish the relationship between the elapsed time T in days since an ocean fish since an ocean sunfish is born and its mass M of T in milligrams is modeled by the following function all right complete the following sentence about the daily percent change in the mass of the sunfish every day there is a blank percent addition or removal from the mass of the sunfish so one thing that we can we know from almost from the get-go we know that the that the sunfish gains weight and we also see that as T grows as T grows the exponent here is going to grow and if you grow an exponent on something that is larger than one M of T is going to grow so I already know it's going to be about addition to the mass of the sunfish but let's think about how much is added every day so let's think about it well let's see if we can rewrite this this is I'm going to just focus on the right hand side of this expression so 1.35 to the T over six plus five that's the same thing as 1.3 five to the fifth power times one point three five to the T over sixth power and that's going to be equal to one point three five to the fifth power times one point three five and I can separate this T over six as one six times T so one point three five to the one sixth power and then that being raised to the teeth to the teeth power so let's think about it every day as T increases by one now we can say that we're going to take the previous days the previous days mass and multiply it by this common ratio the common ratio here isn't the way I've written it isn't one point three five it's one point three five to the one sixth power let me draw a little table here to make that really really clear and all of that algebraic manipulation I just did is just so I could simplify this so I have some common ratio to the teeth power so T and M of T so based on how I've just written it when T is zero well as T is 0 this is 1 so then we just have our initial our initial aim at our initial mass is going to be one point three five to the fifth power and then when T is equal to 1 when T is equal to 1 it's going to be our initial mass one point three five to the fifth power times our common ratio times one point three five to the one sixth power when T equals two we're just going to multiply what we had at T equals one we're just going to multiply that times one point three five to the one sixth again and so every day well let me get every day we are growing every day we're growing by our common ratio one point three five to the one sixth power actually let me get a calculator out we're allowed to use calculators in this exercise so one point three five to the to the open parenthesis 1 divided by six closed parenthesis power is equal to 1.0 1.0 five I'll say one point oh five 1 approximately so this is approximately one point oh five one so we could say this is approximately one point three five times 1.05 one to the teeth power so every day we are growing by a factor of one point oh five one well growing by a factor of one point oh five one means that you're adding a little bit more than five percent you're adding zero point five one every day of your mass so that's you're adding five point one percent and if you're rounding to the nearest percent we would say there is a 5 percent addition to the mass of the sunfish every day