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3 years ago

180/9times10^-5 what’s the answer to this in scientific notation

Mathematics

1 answer:

Lady bird [3.3K]3 years ago

7 0

Answer:

180/9= 20

20x10^-5= 0.00020

Step-by-step explanation:

Move the decimal point 5 times to the left :)

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Answer:

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Step-by-step explanation:

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Look at where the line crosses the X axis.

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3 years ago

For a treasure hunt game 300 balls were hidden of the balls that were hidden 40% of them were yellow and the rest are white Sall

astraxan [27]

Step-by-step explanation:

So first you want to multiply the percentage of the yellow balls by three to get the amount of "300%" which is 300 balls. Munising 100 by 40, you get 60% is white. So multiply that by three also. So for Yellow, you have 120 balls, and for white, you have 180. So you take 120 and take 20% of it, and it would give us 24. You take the 180, and that one gives us 144. So if you add 144 and 24, Sally's team found 168 balls out of 300.

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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>

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

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<h3><u>Explore More</u></h3>

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{sinx}{x} = 1}}$

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$\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{ {a}^{x} - 1}{x} = loga}}$

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