Answer:
50
Step-by-step explanation:
42/n³∑k²+12/n²∑k+30/n∑1
=42/n³[n(n+1)(2n+1)/6]+12/n²[n(n+1)/2]+30/n [n]
=7n(n+1)(2n+1)/n³+6n(n+1)/n²+30
=7(n+1)(2n+1)/n²+6(n+1)/n+30
=[7(2n²+3n+1)+6(n²+n)+30n²]/n²
=[14n²+21n+7+6n²+6n+30n²]/n²
=[50n²+27n+7]/n²
=[50+27/n+7/n²]
→50 as n→∞
because 1/n,1/n²→0 as n→∞
my brother did that too but I'm sorry i cant help you but I'm gonna say just keep up the good work
Answer:
cot(∅) = 
Step-by-step explanation:
tan ∅ and cot ∅ are inverse functions
therefore the inverse of
tan ∅ = √15 / 10
is equal to
cot ∅ = 10 / √15
Rationalizing the denominator
* 
cot ∅ =
I think you mean,
You can use the usual formula,
That in this case is,
Where I have used,
When we Simplify [(x^2)^3 × 5x] / [6x^2 × 15x^3], the result obtained is (1/18)x^2
<h3>Data obtained from the question</h3>
- [(x^2)^3 × 5x] / [6x^2 × 15x^3]
- Simplification =?
<h3>How to simplify [(x^2)^3 × 5x] / [6x^2 × 15x^3]</h3>
[(x^2)^3 × 5x] / [6x^2 × 15x^3]
Recall
(M^a)^b = M^ab
Thus,
(x^2)^3 = x^6
- [(x^2)^3 × 5x] / [6x^2 × 15x^3] = [x^6 × 5x] / [6x^2 × 15x^3]
Recall
M^a × M^b = M^(a+b)
Thus,
x^6 × 5x = 5x^(6 + 1) = 5x^7
6x^2 × 15x^3] = (6×15)x^(2 + 3) = 90x^5
- [x^6 × 5x] / [6x^2 × 15x^3] = 5x^7 / 90x^5
Recall
M^a ÷ M^b = M^(a - b)
Thus,
5x^7 ÷ 90x^5 = (5÷90)x^(7 - 5) = (1/18)x^2
Therefore,
- [(x^2)^3 × 5x] / [6x^2 × 15x^3] = (1/18)x^2
Learn more about algebra:
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