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4 years ago

Give me the answer for 2 1/5 times 4

Mathematics

2 answers:

pav-90 [236]4 years ago

7 0

The answer for 2 1/5 times for total is 8.8

Korvikt [17]4 years ago

6 0

2 1/5* 4
= (11/5) * 4
= 44/5
= (40+ 4)/5
= 40/5+ 4/5
= 8+ 4/5
= 8 4/5

The final answer is 8 4/5~

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How do I solve this equation quick with an easy method?

Pani-rosa [81]

Answer:

4/(3(x-2))

Step-by-step explanation:

3x^2-21x+30=3(x^2-7x+10)=3(x-5)(x-2)

3x-15=3(x-5)

----------------------------

So the common denominator must be 3(x-5)(x-2)

2(x-2)=2x-4

Add the numerators,

(2x-16)+(2x-4)=4x-16-4=4x-20

-----------------

(4x-20)/[3(x-5)(x-2)]

simplify 4x-20 into 4(x-5)

cancel out the (x-5)'s for both the denominator and the numerator

4/[3(x-2)]

3 0

3 years ago

What is the answer to my question? No links please. Urgent.

babunello [35]

Answer:

57cm^3

i belive this is right.

6 0

3 years ago

Read 2 more answers

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icang [17]

The answer is the top right one.

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4 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>

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

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

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8 0

3 years ago

The table shows a proportional relationship.

sergey [27]

.8 is the answer you add .8 to x

4 0

3 years ago