<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

Well depending on how you look at you could do $40 plus that extra 20% and since you said that means to add you would do the same price. as the actual price before you add that thirty 20% so you would add 40 plus 40 equals 80. If you did 25% the you would add 40 and 40 which is 80 and that's 20%out of 25% and add 5 to 80 and you get $85.
Answer: 81.2
Step-by-step explanation:
h(t) = -5t² + 40t + 1.2
h(4) = -5(4)² + 40(4) + 1.2 We plug 4 seconds as t, and solve h(4)
h(4) = -80 + 160 + 1.2
h(4) = 80 + 1.2
h(4) = 81.2
Answer:
1st problem: b) 
2nd problem: c) 
Step-by-step explanation:
1st problem:
The formula/equation you want to use is:

where
t=number of years
A=amount he will owe in t years
P=principal (initial amount)
r=rate
n=number of times the interest is compounded per year t.
We are given:
P=2500
r=12%=.12
n=12 (since there are 12 months in a year and the interest is being compounded per month)

Time to clean up the inside of the ( ).


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2nd Problem:
Compounded continuously problems use base as e.

P is still the principal
r is still the rate
t is still the number of years
A is still the amount.
You are given:
P=2500
r=12%=.12
Let's plug that information in:
.
Answer:
i do not know what the image loos like but it should be D
Step-by-step explanation:
It depicts Spanish horses adapted for use by Native Americans.