Answer:
is it a decimal
Step-by-step explanation:
Answer:
Jade has 9 cars
Step-by-step explanation:
<em>There is no specific requirement in the question, but I'm assuming you need to compute the time needed for Alexis reach 1,000,000 Instagram followers</em>
Answer:

Step-by-step explanation:
<u>Exponential Growth
</u>
When the number of observed elements grows as the previous value multiplied by a constant ratio, we have exponential growth. The formula to model such situations is

Where
is the initial value of f, 1 + r is the constant ratio, and t is the time expressed in half days (12 hours)
The initial value is 100 Instant followers, thus:

We need to know when the number of followers will reach 1,000,000. Setting up the equation

Simplifying by 100

Taking logarithms


Solving for t
periods of 12 hrs

Answer:
-2
Step-by-step explanation:
I think this was the question you were talking about.