Mrs. Cho wrote the following problem on the board. Problem: StartFraction 1 Over x squared EndFraction minus StartFraction 2 Ove
2 answers:
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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
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
Answer:
Step-by-step explanation:
To find the area of the shaded region, you must find the area of the larger rectangle [(x+10)×(2x+5)] and subtract the area of the smaller rectangle [(x+1)×(x+1)] from it.
First, you simplify the larger rectangle's area:
(x+10)×(2x+5) =
Then, simplify the smaller rectangle's area:
(x+1)×(x+1) =
Finally, subtract the smaller rectangle's area from the larger rectangle's area:
Therefore, the final answer is .
I hope this helps!
Answer:
1
Step-by-step explanation:
To find x, set both equations equal to each other.