The question asks: "Let
. Find the largest integer n so that <span>f(2) · f(3) · f(4) · ... · f(n-1) · f(n) < 1.98"
The answer is n = 98</span>Explanation:
First thing, consider that the function can be written as:
Now, let's expand the product, substituting the function with its equation for the requested values:
As you can see, the intermediate terms cancel out with each other, leaving us with:
This is a simple inequality that can be easily solved:
200n < 198(n + 1)
200n < 198n + 198
2n < 198
n < 99
Hence, the greatest integer n < 99 (extremity excluded) is
98.
Answer:
Step-by-step explanation:
Use KCF method for fraction division.
Keep the first fraction
Change the operation to multiplication
Flip the second fraction.
1) 3÷ (1/4) =
2) 7 ÷ (1/2) =
3) (1/5) ÷ (1/20) =
4) (1/2) ÷ (1/10) =
5) (1/7) ÷ 3 =
Answer:
It would be D, 1/4.
Step-by-step explanation:
I hope this helps you!
Add 12 to both sides. leaving you with
3x=21
then divide 3 on both sides. leaving you with
x=7
hopefully that’s right!!!
Answer: 2,3
Step-by-step explanation:
just true statements. Arithmetic reasoning.