The value of -0-6- is -approximate upto three decimal-0-7350-7750-7550-745

The correct option is B 0.775Let y nbsp; = √(x) ⇒dy/dx = 1/2 √(x) Δ y = √(x + Δ x) √(x) Putting x = 0.64 and Δ x = 0.04 ⇒Δ y = √(0.6) √(0.64) ⇒ nbsp; ...

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Standard XII

Mathematics

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The value of ...

Question

The value of $\sqrt{0.6}$ is (approximate upto three decimal)

0.735

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0.745

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0.775

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0.755

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Solution

The correct option is C $0.775$
Let $y = \sqrt{x}$
$\Rightarrow \frac{d y}{d x} = \frac{1}{2 \sqrt{x}}$
$Δ y = \sqrt{x + Δ x} - \sqrt{x}$
Putting $x = 0.64$ and $Δ x = - 0.04$
$\Rightarrow Δ y = \sqrt{0.6} - \sqrt{0.64} \Rightarrow \sqrt{0.6} = Δ y + 0.8$
Now $Δ y = \frac{d y}{d x} \times Δ x = \frac{Δ x}{2 \sqrt{x}}$
$\Rightarrow Δ = - \frac{0.04}{2 \times 0.8} = - 0.025$
So $\sqrt{0.6} = - 0.025 + 0.8 = 0.775$

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Using differentials, find the approximate value of the following up to 3 places of decimal.
$\sqrt{0.6}$

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The value of √0.6 is (approximate upto three decimal)

The correct option is C 0.775Let y=√x ⇒dydx=12√x Δy=√x+Δx−√x Putting x=0.64 and Δx=−0.04 ⇒Δy=√0.6−√0.64⇒√0.6=Δy+0.8 Now Δy=dydx×Δx=Δx2√x ⇒Δ=−0.042×0.8=−0.025 So √0.6=−0.025+0.8=0.775

The value of $\sqrt{0.6}$ is (approximate upto three decimal)