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Expansion of (x^2 - y^2)

The value of ...

Question

$The value of (0.9)^{2} - (0.6)^{2} is$

0.54

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- 0.54

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- 0.45

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0.45

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Solution

The correct option is D
0.45

$(0.9)^{2} - (0.6)^{2} = (0.9 + 0.6) (0.9 - 0.6)$
$= (1.5) (0.3)$
$= 0.45$

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Similar questions

The value of $(0.9)^{2} - (0.6)^{2}$ is

E_{1} = {x \in R : x \neq 1 and \frac{x}{x - 1} > 0}

E_{2} = {x \in E_{1} : {sin}^{- 1} ({log}_{e} (\frac{x}{x - 1})) is a real number} .

(

{sin}^{- 1} x

[- \frac{π}{2}, \frac{π}{2}]

)

f : E_{1} \to R

f (x) = {log}_{e} (\frac{x}{x - 1})

g : E_{2} \to R

g (x) = {sin}^{- 1} ({log}_{e} (\frac{x}{x - 1})) .

\begin{matrix} L i s t A & L i s t B P . & The range of f is & 1. & (- \infty, \frac{1}{1 - e}] \cup [\frac{e}{e - 1}, \infty) Q . & The range of g contains & 2. & (0, 1) R . & The domain of f contains & 3. & [- \frac{1}{2}, \frac{1}{2}] S . & The domain of g is & 4. & (- \infty, 0) \cup (0, \infty) 5. & (- \infty, \frac{e}{e - 1}] 6. & (- \infty, 0) \cup (\frac{1}{2}, \frac{e}{e - 1}] \end{matrix}

In the figure given, AOB is a straight line. What is

the value of a

The value of ${0.6}^{2}$ is

List I

List II

List I

List II .

\begin{matrix} List I & List II (A) & Let f be a real-valued differentiable & (P) & 18 function on R such that f^{'} (1) = 6 and f^{'} (2) = 2. Then lim h \to 0 \frac{f (3 cos h + 4 sin h - 2) - f (1)}{f (3 e^{h} - 5 sec h + 4) - f (2)} is equal to (B) & If (0.5)^{{log}_{3} {log}_{1 / 5} (x^{2} - \frac{4}{5})} > 1, then the least & (Q) & 4 positive integral value of x is (C) & The value of lim x \to 1 \frac{{sin}^{2} (x^{3} + x^{2} + x - 3)}{1 - cos (x^{2} - 4 x + 3)} is & (R) & 5 (D) & Number of points of non-differentiability of & (S) & 2 f (x) = min (| sin x |, | cos x |, \frac{1}{4}) in (0, π) is \end{matrix}

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