The problem can be solved step by step, if we know certain basic rules of summation. Following rules assume summation limits are identical.




Armed with the above rules, we can split up the summation into simple terms:





=> (a)
f(x)=28n-n^2=> f'(x)=28-2n
=> at f'(x)=0 => x=14
Since f''(x)=-2 <0 therefore f(14) is a maximum
(b)
f(x) is a maximum when n=14
(c)
the maximum value of f(x) is f(14)=196
The domain of the function is the set of all x's that are suitable for the given equation. In the problem, we are asked to determine the equation with the most restricted domain. Option A can have x from negative infinity to positive infinity as well as option B and option D. OPtion C can only have x equal to zero and all positive numbers. Hence the answer is C. hope that helped
480 + 192 = 672 students total
16 + 12 = 28 classrooms total
672/28 = 24 students per room
24 * 12 = 288 students that need to be in school B
288 - 192 = 96 students who have to transfer
Answer:
a/ (5r^2) = t
Step-by-step explanation:
a= 5r^2 t
Divide each side by 5r^2
a/ 5r^2 = 5r^2 t/5r^2 t
a/ 5r^2 = t