The initial value of 100 that doubles over each interval.
without the answer choices, I can only describe it and give you an example of the graph.
I'm assuming the function is 100*(2)^x because if it is as listed it would be a quadratic function with a vertical stretch of 100.
Answer:
The smallest positive integer solution to the given system of congruences is 30.
Step-by-step explanation:
The given system of congruences is
where, m and n are positive integers.
It means, if the number divided by 5, then remainder is 0 and if the same number is divided by 11, then the remainder is 8. It can be defined as
Now, we can say that m>n because m and n are positive integers.
For n=1,
19 is not divisible by 5 so m is not an integer for n=1.
For n=2,
The value of m is 6 and the value of n is 2. So the smallest positive integer solution to the given system of congruences is
Therefore the smallest positive integer solution to the given system of congruences is 30.
Answer:
Step-by-step explanation:
Well we can simplify the numerator, by multiplying the 4 by the 6 and the m^3 and m^4 (add the exponents, explained in one of my previous answers I think)
This gives us the fraction: 
We can now divide the m^7 by m^2 by subtracting the exponents, and the reason why this works, is you're simply cancelling out the m's, If we express this in expanded form we have the following fraction: 
Since there is two m's in the denominator and there is also two (more than two) m's in the numerator, we can cancel those two m's out, and we get the fraction:
which can be simplified in exponent form as: 
, now all we have to do is divide the 24 by the 3, to get 8
This gives us the answer:
