Factorise: $(0.7)^{2} - (0.3)^{2}$

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Solution

Given expression is $(0.7)^{2} - (0.3)^{2}$

We know that, $a^{2} - b^{2} = (a + b) (a - b)$

Here, $a = 0.7$ and $b = 0.3$

$∴ (0.7)^{2} - (0.3)^{2} = (0.7 + 0.3) (0.4 - 0.3)$

Hence, $(0.7)^{2} - (0.3)^{2} = (0.7 + 0.3) (0.4 - 0.3)$ .

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Similar questions

Q. Factorise by using formula:

(0.7)^{2} - (0.3)^{2}

is 0.4
If true then enter

1

and if false then enter

0

$(0.7)^{2} - (0.3)^{2}$

Q. The difference

(0.7)^{2} - (0.3)^{2}

simplifies to

Factorise:

(i) $4 a^{2} - 25$ (ii) $x^{2} - \frac{9}{16}$ (iii) $x^{4} - y^{4}$ (iv) $[7 \frac{3}{10}]^{2} - [2 \frac{1}{10}]^{2}$ (v) $(0.7)^{2} - (0.3)^{2}$ (vi) $(5 a - 2 b)^{2} - (2 a - b)^{2}$

Q. In the reaction

4 A + 2 B + 3 C \to A_{4} B_{2} C_{3}

The number of moles of product formed, starting from one mole of A, 0.6 mole of B and 0.72 mole of C?

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Factorise: (0.7)2−(0.3)2

Given expression is (0.7)2−(0.3)2We know that, a2−b2=(a+b)(a−b)Here, a=0.7 and b=0.3∴ (0.7)2−(0.3)2=(0.7+0.3)(0.4−0.3)Hence, (0.7)2−(0.3)2=(0.7+0.3)(0.4−0.3).

Factorise: $(0.7)^{2} - (0.3)^{2}$

Given expression is $(0.7)^{2} - (0.3)^{2}$

We know that, $a^{2} - b^{2} = (a + b) (a - b)$

Here, $a = 0.7$ and $b = 0.3$

$∴ (0.7)^{2} - (0.3)^{2} = (0.7 + 0.3) (0.4 - 0.3)$

Hence, $(0.7)^{2} - (0.3)^{2} = (0.7 + 0.3) (0.4 - 0.3)$ .