2000=1800(1.04)^t,,log(1.11)/log(1.04)=t use the calculator,,,hope it helps :-)
The range of the equation is 
Explanation:
The given equation is 
We need to determine the range of the equation.
<u>Range:</u>
The range of the function is the set of all dependent y - values for which the function is well defined.
Let us simplify the equation.
Thus, we have;

This can be written as 
Now, we shall determine the range.
Let us interchange the variables x and y.
Thus, we have;

Solving for y, we get;

Applying the log rule, if f(x) = g(x) then
, then, we get;

Simplifying, we get;

Dividing both sides by
, we have;

Subtracting 7 from both sides of the equation, we have;

Dividing both sides by 2, we get;

Let us find the positive values for logs.
Thus, we have,;


The function domain is 
By combining the intervals, the range becomes 
Hence, the range of the equation is 
What grade are u? I'm just asking cause the formulas are going with grade I could help if u want me too
Answer:
U it is DG if jug chuchi DG u want u back to the u and composition
Answer:
✅On average Albany gets 20.967 inches rain pee year and Zachary gets 25.153 inches rain per year.
✅On average, it rains more in Zachary in a year.
Step-by-step explanation:
Mean annual rainfall on average = total amount of rainfall for the period of years ÷ 10
✔️Mean Annual rainfall for Albany (in.):
= (22.82 + 19.83 + 28.40 + 23.35 + 30.78 + 21.34 + 11.20 + 17.04 + 18.63 + 16.28) ÷ 10
= 20.967
✔️Mean Annual rainfall for Zachary (in.):
= (26.41 + 23.54 + 30.02 + 26.35 + 34.24 + 25.74 + 18.98 + 21.26 + 23.61 + 21.38) ÷ 10
= 25.153
✅Therefore, on average Albany gets 20.967 inches rain pee year and Zachary gets 25.153 inches rain per year.
✅On average, it rains more in Zachary in a year.