The query is a mathematical problem about diophantine equations and the Chinese Remainder Theorem. By setting up the equations 32n + 30 and 58m + 44, we search for a number that fits both conditions and is less than 3000. That number is 1978.
The problem described is a common type of question in number theory, specifically in the field of diophantine equations. In mathematics, a diophantine equation is a polynomial equation where the solutions are sought in integers. This problem consists in finding a common remainder when dividing by different numbers, which is the essence of the Chinese Remainder Theorem.
We can set up the equations as follows: the number can be written as 32n + 30 (this gives a remainder of 30 when divided by 32) and as 58m + 44 for some integers n and m (this gives a remainder of 44 when divided by 58). Now, we check for possible solutions less than 3000 by trying out different values of 'n' and 'm'.
After checking several possibilities one by one, the smallest positive number that satisfies both equations is 1978.
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Answer:
f(2)=128
Step-by-step explanation:
f(x) = 2(8)^x evaluate f(2)
f(2)= 2(8)^2
f(2)=2(64)
f(2)=128
Answer:
30 in^3
Step-by-step explanation:
Divide 159 by 5. The volume of the jar is 30 in^3
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