the is c.32$&$&#&#
Step-by-step explanation:
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Answer: 1, 5, 2
All you do is subtract 3 from the y
Answer:
Step-by-step explanation
if my answer is wrong i'm sorry here is a manual on how to do it if its wrong
i think they are right but just in case
Identify the output values. If each input value leads to only one output value, classify the relationship as a function. If any input value leads to two or more outputs, do not classify the relationship as a function.
1 relation
2 function
3 relation
4 function
i know 4 is right but i don't know about others double check i suggest just in case i think they are right
![\rm 5=e^{3b}](https://tex.z-dn.net/?f=%5Crm%205%3De%5E%7B3b%7D)
The unknown b is stuck in the exponent position.
We can can fix that by using logarithms.
Log is the inverse operation of the exponential.
We'll take log of each side.
Log of what base tho?
Well, the base of our exponential is e,
so we'll take log base e of each side.
![\rm log_e(5)=log_e(e^{3b})](https://tex.z-dn.net/?f=%5Crm%20log_e%285%29%3Dlog_e%28e%5E%7B3b%7D%29)
We'll apply one of our log rules next:
![\rm \log(x^y)=y\cdot\log(x)](https://tex.z-dn.net/?f=%5Crm%20%5Clog%28x%5Ey%29%3Dy%5Ccdot%5Clog%28x%29)
This allows us to take the exponent out of the log,
![\rm log_e(5)=(3b)log_e(e)](https://tex.z-dn.net/?f=%5Crm%20log_e%285%29%3D%283b%29log_e%28e%29)
Another thing to remember about logs:
When the base of the log matches the inside of the log,
then the whole thing is simply 1,
![\rm log_{10}(10)=1](https://tex.z-dn.net/?f=%5Crm%20log_%7B10%7D%2810%29%3D1)
![\rm log_5(5)=1](https://tex.z-dn.net/?f=%5Crm%20log_5%285%29%3D1)
![\rm log_e(e)=1](https://tex.z-dn.net/?f=%5Crm%20log_e%28e%29%3D1)
So our equation simplifies to this,
![\rm log_e(5)=(3b)\cdot1](https://tex.z-dn.net/?f=%5Crm%20log_e%285%29%3D%283b%29%5Ccdot1)
As a final step, divide both sides by 3,
![\rm \frac13log_e(5)=b](https://tex.z-dn.net/?f=%5Crm%20%5Cfrac13log_e%285%29%3Db)
k, hope that helps!