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2 years ago

Find the domain and the range of {(1,7),(3,8),(0,4),(-1,4)}

Mathematics

1 answer:

jenyasd209 [6]2 years ago

6 0

Answer:

domain={1,3,0,-1}

range={7,8,4}

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Determine whether the function is periodic. If it is periodic, find the period.

puteri [66]

Answer:

none of them

$f(x) = a \: \sin(b + cx) \\ t = \frac{2\pi}{ |c| } \\ t = \frac{2\pi}{ |1| } = 2\pi$

this is the right answer

6 0

3 years ago

Lim n-> infinity [1/3 + 1/3² + 1/3³ + . . . .+ 1/3ⁿ]

Verizon [17]

Answer:

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\:\displaystyle\lim_{n \to \infty }\rm \bigg[\dfrac{1}{3} + \dfrac{1}{ {3}^{2} } + \dfrac{1}{ {3}^{3} } + - - + \dfrac{1}{ {3}^{n} } \bigg]$

Let we first evaluate

$\rm :\longmapsto\:\dfrac{1}{3} + \dfrac{1}{ {3}^{2} } + \dfrac{1}{ {3}^{3} } + - - + \dfrac{1}{ {3}^{n} }$

Its a Geometric progression with

$\rm :\longmapsto\:a = \dfrac{1}{3}$

$\rm :\longmapsto\:r = \dfrac{1}{3}$

$\rm :\longmapsto\:n = n$

So, Sum of n terms of GP series is

$\rm :\longmapsto\:S_n = \dfrac{a(1 - {r}^{n} )}{1 - r}$

$\rm :\longmapsto\:S_n = \dfrac{1}{3} \bigg[\dfrac{1 - {\bigg[\dfrac{1}{3} \bigg]}^{n} }{1 - \dfrac{1}{3} } \bigg]$

$\rm :\longmapsto\:S_n = \dfrac{1}{3} \bigg[\dfrac{1 - {\bigg[\dfrac{1}{3} \bigg]}^{n} }{\dfrac{3 - 1}{3} } \bigg]$

$\rm :\longmapsto\:S_n = \dfrac{1}{3} \bigg[\dfrac{1 - {\bigg[\dfrac{1}{3} \bigg]}^{n} }{\dfrac{2}{3} } \bigg]$

$\bf\implies \:S_n = \dfrac{1}{2}\bigg[1 - \dfrac{1}{ {3}^{n} } \bigg]$

<u>Hence, </u>

$\bf :\longmapsto\:\dfrac{1}{3} + \dfrac{1}{ {3}^{2} } + \dfrac{1}{ {3}^{3} } + - - + \dfrac{1}{ {3}^{n} } = \dfrac{1}{2}\bigg[1 - \dfrac{1}{ {3}^{n} } \bigg]$

<u>Therefore, </u>

$\purple{\rm :\longmapsto\:\displaystyle\lim_{n \to \infty }\rm \bigg[\dfrac{1}{3} + \dfrac{1}{ {3}^{2} } + \dfrac{1}{ {3}^{3} } + - - + \dfrac{1}{ {3}^{n} } \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty }\rm \dfrac{1}{2}\bigg[1 - \dfrac{1}{ {3}^{n} } \bigg]$

$\rm \: = \: \rm \dfrac{1}{2}\bigg[1 - 0 \bigg]$

$\rm \: = \: \rm \dfrac{1}{2}$

<u>Hence, </u>

$\purple{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{n \to \infty }\rm \bigg[\dfrac{1}{3} + \dfrac{1}{ {3}^{2} } + \dfrac{1}{ {3}^{3} } + - - + \dfrac{1}{ {3}^{n} } \bigg]} = \frac{1}{2}}}$