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3241004551 [841]
3 years ago
7

If n is a positive integer, then lim (n-->infinity) (1/n) [1/(1+(1/n)) + 1/(1+(2/n))+...+1/(1+(n/n)] is?

Mathematics
1 answer:
Anton [14]3 years ago
3 0
The correct answer to this question is <span>d.) integral from 1 to 2 of (2/(x+1))
</span>To solve this:

Since Δx = 1/n. 
lim (n→∞) Δx [1/(1+Δx) + 1/(1+2Δx)+ ... + 1/(1+nΔx)] 
= lim (n→∞) Σ(k = 1 to n) [1/(1 + kΔx)] Δx. 

x <---> a + kΔx 

a = 0, then b = 1, so that Δx = (b - a)/n = 1/n

Since (1 + kΔx) combination, a = 1 so that b = 2. 
Then, f(1 + kΔx) <-----> f(x) ==> f(x) = 1/x. 

This sum represents the integral 
∫(x = 1 to 2) (1/x) dx, so the correct answer is <span>d.) integral from 1 to 2 of (2/(x+1))

Thank you for posting your question. I hope that this answer helped you. Let me know if you need more help. 
</span>
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<h3>Motion of the golf ball</h3>

The path of motion of the golf ball can be described as parabolic because it is projectile.

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<h3><u>Answer:</u></h3>

\boxed{\pink{\sf \leadsto Yes \ there \ is \ a \ solution \ of \ the \ given \ inequality .}}

<h3><u>Step-by-step explanation:</u></h3>

A inequality is given to us and we need to convert it into standard form and see whether if it has a solution . So let's solve the inequality.

The inequality given to us is :-

\bf\implies |2y + 3 | - 1 \leq 0 \\\\\bf\implies |2y+3|\leq 1 \\\\\bf\implies (|2y+3|)^2 \leq 1^2  \\\\\bf\implies (2y+3)^2 \leq 1  \\\\\bf\implies (2y)^2+3^2+2(2y)(3) \leq 1  \\\\\bf\implies 4y^2+9+12y - 1 \leq 0  \\\\\bf\implies 4y^2+12y+8 \leq 0 \\\\\bf\implies 4( y^2 + 3y + 2 ) \leq 0  \\\\\bf\implies y^2+3y +2 \leq 0 \:\:\bigg\lgroup \purple{\bf Standard \ form \ of \ inequality }\bigg\rgroup   \\\\\bf\implies y^2y+2y+y+2 \leq 0  \\\\\bf\implies y(y+2)+1(y+2)\leq 0  \\\\\bf\implies ( y+2)(y+1)\leq 0  \\\\\bf\implies \boxed{\red{\bf y \leq (-2) , (-1) }}

Let's plot a graph to see its interval . Graph attached in attachment .

Now we can see that the Interval notation of would be ,

\boxed{\boxed{\orange \tt \purple{\leadsto}y \in [-2,-1] }}

<h3><u>Hence</u><u> the</u><u> </u><u>standa</u><u>rd</u><u> </u><u>form</u><u> </u><u>of</u><u> </u><u>inequa</u><u>lity</u><u> </u><u>is</u><u> </u><u>y²</u><u>+</u><u>3y</u><u> </u><u>+</u><u>2</u><u> </u><u>≤</u><u> </u><u>0</u><u> </u><u>and</u><u> </u><u>the </u><u>Solution</u><u> </u><u>set</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>ineq</u><u>uality</u><u> </u><u>is</u><u> </u><u>[</u><u> </u><u>-</u><u>2</u><u> </u><u>,</u><u> </u><u>-</u><u>1</u><u> </u><u>]</u><u> </u><u>.</u></h3>
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