Question 1

i watched this so many times and still dont get none of it

Accepted Answer

you see 1.35^t/6+5=1.35^5times 1.35^t/6, this is logical, because 1.35^5 times 1.35^t/6=1.35^t/6+5
and (1.35^1/6）^t is also logical, bacause he moved the t outside the parentheses, and so the answer is the same, for example if t=5, 5/6= 0.83, and 1/6 times 5/1=5/6,so this is also logical,
but I don't really understand the last step

Question 2

At 3:51, I repeated the last part many times yet I still don't get it, I just don't understand how 5% is the answer

Accepted Answer

The common ratio is 1.051, which means that every time t is increased by 1, M(t) is multiplied by 1.051. Since 1.051 is larger than 0, M(t) will be increasing. 1.051 = 105.1%, and since anything multiplied by 100% is itself, M(t) will increase by 5.1% (105.1-100) every day. Hope this helps!

Question 3

can someone please explain this video in simpler terms? like a step-by-step process? thank you.

Accepted Answer

So basically what we want to find in the video is the common ratio for 1 day but we are given the common ratio for t/5 +6 days. By using the properties of exponents. we already know, Sal simplifies it in a way such that the power of the common ratio is 1. Hope you find this helpful.

Question 4

Does this mean that the initial mass of the sunfish is 1.35^5 = 4.48...?

Accepted Answer

Yes.  It t=0 (no time since birth), then the sunfish weighs 1.35^5

Question 5

Can someone give me a few examples of how to solve problems like this? Im kinda-very stuck on it...

Thankyou!🙂

Accepted Answer

Here is an example taken from the exercise following this video:
A sample of an unknown chemical element naturally loses its mass over time.
The relationship between the elapsed time t, in days, since the mass of the sample was initially measured, and its mass, M(t), in grams, is modeled by the following function:
M(t)=900⋅(8/27)^t
The sample loses 1/3 of its mass every ______ days.
Let's look at the different parts of the equation.
900 grams is the sample's initial mass.
Multiplying that by (8/27)^t means that by the end of each day only 8/27 of the amount that was there at the beginning of that day is left. In other words, the sample loses 19/27 of its mass over the course of one day. This is more than 1/3, so it must be that the sample loses 1/3 of its mass in less than one day.
For the purpose of this exercise, we can mostly ignore the original mass, 900.
If we can change the common ratio from 8/27 to 2/3 somehow, then we will be able to find out how long it takes for the sample to lose 1/3 of its mass.
8/27=2/3^3, so:
(8/27)^t=
((2/3)^3)^t=
(2/3)^3t
This means that after about 1/3 of a day 2/3 of the mass that was there at the beginning of the day remains. In other words, the sample loses 1/3 of its mass approximately every 0.33 days.
Hopefully that was helpful. If there is anything unclear or anything you don't understand about this, then you can ask about it.

Question 6

Could someone please explain why my strategy didn't work? I made a table with values for t (0 days, 6 days, 12 days) which made it clear that every six days we multiply the mass by 1.35, which is the same thing as adding 35%. Then I thought that if we add 35% every six days, then if we divide 35 by 6 we'll get the percent we add every day. But then I got approximately 5.8% as a result instead of 5.1%

Accepted Answer

Your strategy assumes we add the same amount every day, which we don't (that would be linear growth, not exponential). You computed the _average_ amount added per day over those six days, but because the growth is exponential, we have less growth in the beginning and more growth later.

Question 7

How and why does the expression have an initial value of 1.35^5? If the common ratio of the growth factor would be 1.051^t and it is growing since birth, where did the initial value come from?

Accepted Answer

The expression above is said to model its mass "since an ocean sunfish is born". Apparently, the fish is born with some positive mass, which is the value of the expression at the time t = 0, i.e. 1.35^5 kilos (or grams, or some other unit of weight). Representing an initial value under a common exponent with a growth factor is just a trick to make the expression more compact (and, actually, more sophisticated). This is generally doable, as any positive number (initial value) may be represented as some exponent of any other positive number (growth factor). In this particular case, the initial value is first represented as a 5th power of the growth factor and then they two are blended under the single exponent of (t/6+5).

Algebra 2 (Eureka Math/EngageNY)

Course: Algebra 2 (Eureka Math/EngageNY) > Unit 3

Interpreting change in exponential models: with manipulation

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