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3 years ago

Sam have 1 1/5 meters of rope he cut off 1/2 meters and use it for a project for school how much real does Sam have left

Mathematics

2 answers:

Yuri [45]3 years ago

7 0

Answer is 7/10

First you find the common denominator which is 10 from doing 5*2.

Then you do 1*2 to get 2 and then also 5*1 to get 5.

Now you have 5/10 - 2/10 = 3/10.

Then you combine the whole and fraction parts and get..

1 - 3/10 = 7/10

( 10-3=7)

BartSMP [9]3 years ago

4 0

Hope this helps also has the decimal

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Divide x^3 -3x^2 by x-2 . Step 1 - Fill in the missing number:

vagabundo [1.1K]

Answer:

Step-by-step explanation:

First let us write the given polynomial as in descending powers of x with 0 coefficients for missing items

F(x) = x^3-3x^2+0x+0

We have to divide this by x-2

Leading terms in the dividend and divisor are

x^3 and x

Hence quotient I term would be x^3/x=x^2

x-2) x^3-3x^2+0x+0(x^2

x^3-2x^2

Multiply x-2 by x square and write below the term and subtract

We get

x-2) x^3-3x^2+0x+0(x^2

x^3-2x^2

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-x^2+0x

Again take the leading terms and find quotient is –x

x-2) x^3-3x^2+0x+0(x^2-x

x^3-2x^2

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

-x^2+0x

-x^2-2x

Subtract to get 2x +0 as remainder.

x-2) x^3-3x^2+0x+0(x^2-x-2

x^3-2x^2

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-x^2+0x

-x^2+2x

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-2x-0

-2x+4

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-4

Thus remainder is -4 and quotient is x^2-x-2

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tankabanditka [31]

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Step-by-step explanation:

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saul85 [17]

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3 years ago

D/d{cosec^-1(1+x²/2x)} is equal to

SIZIF [17.4K]

Step-by-step explanation:

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

$\rm :\longmapsto\:\dfrac{d}{dx} {cosec}^{ - 1} \bigg( \dfrac{1 + {x}^{2} }{2x} \bigg)$

Let assume that

$\rm :\longmapsto\:y = {cosec}^{ - 1} \bigg( \dfrac{1 + {x}^{2} }{2x} \bigg)$

We know,

$\boxed{\tt{ {cosec}^{ - 1}x = {sin}^{ - 1}\bigg( \dfrac{1}{x} \bigg)}}$

So, using this, we get

$\rm :\longmapsto\:y = sin^{ - 1} \bigg( \dfrac{2x}{1 + {x}^{2} } \bigg)$

Now, we use Method of Substitution, So we substitute

$\red{\rm :\longmapsto\:x = tanz \: \rm\implies \:z = {tan}^{ - 1}x}$

So, above expression can be rewritten as

$\rm :\longmapsto\:y = sin^{ - 1} \bigg( \dfrac{2tanz}{1 + {tan}^{2} z} \bigg)$

$\rm :\longmapsto\:y = sin^{ - 1} \bigg( sin2z \bigg)$

$\rm\implies \:y = 2z$

$\bf\implies \:y = 2 {tan}^{ - 1}x$

So,

$\bf\implies \: {cosec}^{ - 1}\bigg( \dfrac{1 + {x}^{2} }{2x} \bigg) = 2 {tan}^{ - 1}x$

Thus,

$\rm :\longmapsto\:\dfrac{d}{dx} {cosec}^{ - 1} \bigg( \dfrac{1 + {x}^{2} }{2x} \bigg)$

$\rm \: = \: \dfrac{d}{dx}(2 {tan}^{ - 1}x)$

$\rm \: = \: 2 \: \dfrac{d}{dx}( {tan}^{ - 1}x)$

$\rm \: = \: 2 \times \dfrac{1}{1 + {x}^{2} }$

$\rm \: = \: \dfrac{2}{1 + {x}^{2} }$

Hence, 

$\purple{\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx} {cosec}^{ - 1} \bigg( \dfrac{1 + {x}^{2} }{2x} \bigg) = \frac{2}{1 + {x}^{2} }}}}$

Hence, Option (d) is correct.

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2 years ago