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zimovet [89]
3 years ago
11

What is the fifth term of a sequence whose first term is 5 and whose common ratio is 3?

Mathematics
2 answers:
Softa [21]3 years ago
7 0

The answer is B, 405.


Deffense [45]3 years ago
6 0
<span>a=first term
r=common ratio
n=the term 
a(5)=5*3(5-1)
a(5)=5*81
a(5)=405
The answer is B)</span>
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2 years ago
Limit (sin4x-4sinx)/x^3 when x close to zero
BartSMP [9]

\Large \boxed{\sf \bf \ \ \lim_{x\rightarrow0} \ {\dfrac{sin(4x)-4sin(x)}{x^3}}=-10 \ \  }

Step-by-step explanation:

Hello, please consider the following.

Using Maclaurin series expansion, we can find an equivalent of sin(x) in the neighbourhood of 0.

sin(x) \sim  \left(x-\dfrac{x^3}{3!}\right)\\\\\text{So, in the neighbourhood of 0}\\\\\begin{aligned}(sin(4x)-4sin(x)) &\sim \left( 4x-\dfrac{(4x)^3}{3!}-4x+\dfrac{4x^3}{3!}\right)\\\\&\sim \left(\dfrac{x^3*4*(1-4^2)}{3*2}\right)\\\\&\sim \left(\dfrac{x^3*2*(-15)}{3}\right)\\\\&\sim \left(x^3*2*(-5)\right)\\\\&\sim \left(x^3*(-10)\right)\\\end{aligned}

Then,

\displaystyle \lim_{x\rightarrow0} \ {\dfrac{sin(4x)-4sin(x)}{x^3}}\\\\= \lim_{x\rightarrow0} \ {\dfrac{-10*x^3}{x^3}}\\\\=-10

Thank you

4 0
3 years ago
Read 2 more answers
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