Question

factorise (0.99)2 using identity
plz plz no square root
it's simply 2 ​

Answer 1

According to Ques.:

(0.99)²

or we can write it as,

(1 - 0.01)²

now by using Identity

(a-b)² = a² + b² -2ab

now applying,

(1-0.01)² = (1)² + (0.01)² - 2× 1× 0.01

= 1+ 0.0001 - 0.02

= 0.9801

hope it help ❣

Answer 2

$(0.99) {2}$

$(1 - 0.01) {}^{2}$

$(a - b) {}^{2} = a {}^{2} + b {}^{2} - 2ab$

$(1 - 0.01) {}^{2} = (1) {}^{2} + (0.01) {}^{2} - 2 \times 1 \times 0.01$

$= 1 + 0.0001 - 0.02$

$0.9801$