Question

please do fast please​

Answer 1

Answer:

The number 104 can be shown as 100+4.

Then the square(104²) can be shown as (100+4)²

$\sf{(100+4)^2$

$\sf{=100^2+2\times4\times100+4^2$

$\sf{=10000+800+16$

$\sf{=10816}$

The number 91 can be shown as 90+1.

Then the square(91²) can be shown as (90+1)²

$\sf{(90+1)^2}$

$\sf{=90^2+2\times90\times1+1^2}$

$\sf{=81000+180+1}$

$\sf{=81181}$

So the square of 104 & 91 is 10816 & 81181 respectively.

More information:

The main idea is the algebraic identity. Here we used the identity $\sf{(a+b)^2=a^2+2ab+b^2}$, which we can substitute any numbers to satisfy the equality.

Identities are equalities true always. You can prove identities with algebra. For example, here we have $\sf{(a+b)^2=a^2+2ab+b^2}$.

$\sf{(a+b)^2}$

$\sf{=(a+b)(a+b)}$

$\sf{=a(a+b)+b(a+b)}$

$\sf{=a^2+ab+ab+b^2}$

$\sf{=a^2+2ab+b^2}$ Hence proven.

Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ 1:

• Find The Square of 104

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♣ ꜰᴏʀᴍᴜʟᴀ ᴜꜱᴇᴅ :

• (a + b)² = a² + b² + 2ab

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♣ ᴀɴꜱᴡᴇʀ :

104 can be written as 100 + 4

(104)² = (100 + 4)²

(104)² = 100² + 4² + 2 × 100 × 4   [Using (a + b)² = a² + b² + 2ab)]

(104)² = 10000 + 16 + 800

(104)² = 10816

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♣ Qᴜᴇꜱᴛɪᴏɴ 2:

• Find The Square of 91 without actual multiplication

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♣ ꜰᴏʀᴍᴜʟᴀ ᴜꜱᴇᴅ :

• (a + b)² = a² + b² + 2ab

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♣ ᴀɴꜱᴡᴇʀ :

91 can be written as 90 + 1

(91)² = (90 + 1)²

(91)² = 90² + 1² + 2 × 90 × 1   [Using (a + b)² = a² + b² + 2ab)]

(91²) = 8100 + 1 + 180

(91)² = 8281