The square of 40 1-4 is

The correct option is D 1620 1/16 (401/4 )^2 can be written as (40 + 1/4 )^2 = (40)^2 + (1/4 )^2 + 2 × 40 ×1/4 [(x+y)^2 = x^2 + y^2 + 2xy ] = 1600 + 1/16 ...

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Byju's Answer

Standard VIII

Mathematics

Deducing Formula for (x+y)^2

The square of...

Question

The square of $40 \frac{1}{4}$ is

$810 \frac{1}{16}$

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$1620 \frac{1}{4}$

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$1620 \frac{1}{8}$

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$1620 \frac{1}{16}$

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Solution

The correct option is D
$1620 \frac{1}{16}$

${(40 \frac{1}{4})}^{2}$ can be written as

${(40 + \frac{1}{4})}^{2}$

$= (40)^{2} + {(\frac{1}{4})}^{2} + 2 \times 40 \times \frac{1}{4} [(x + y)^{2} = x^{2} + y^{2} + 2 x y]$

$= 1600 + \frac{1}{16} + 20$

$= 1620 \frac{1}{16}$

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Similar questions

Q. Let

a = (4^{1 / 401} - 1)

and for each

n \geq 2,

let

b_{n} =^{n} C_{1} +^{n} C_{2} \cdot a +^{n} C_{3} \cdot a^{2} +^{n} C_{n} \cdot a^{n - 1} .

Then the value of

b_{2020} - b_{2019}

0.874 times 401 is:

Q. Using differentials the approximate value of

\sqrt{401}

41 %

of 401 - 250.17 = ? -

77 %

of 910

^{404} C_{4} -^{4} C_{1}^{303} C_{4} +^{4} C_{2}^{202} C_{4} -^{4} C_{3}^{101} C_{4}

is equal to

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The square of 4014 is

The correct option is D 1620116 (4014)2 can be written as (40+14)2 =(40)2+(14)2+2×40×14 [(x+y)2=x2+y2+2xy] =1600+116+20 =1620116

The square of $40 \frac{1}{4}$ is

The correct option is D
$1620 \frac{1}{16}$

${(40 \frac{1}{4})}^{2}$ can be written as

${(40 + \frac{1}{4})}^{2}$

$= (40)^{2} + {(\frac{1}{4})}^{2} + 2 \times 40 \times \frac{1}{4} [(x + y)^{2} = x^{2} + y^{2} + 2 x y]$

$= 1600 + \frac{1}{16} + 20$

$= 1620 \frac{1}{16}$