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3 years ago

Multiple by moving the decimal point 24.6x100,37x1,000 0.367x10,000,0.005x1000,000 .

Mathematics

1 answer:

lesya [120]3 years ago

8 0

The answers are 2460, 37,000, 3670 and 5000

Step-by-step explanation:

Step 1: Find the answers for each by multiplying.

24.6 × 100 = 2460 (moved one decimal point)

37 × 1000 = 37,000

0.367 × 10000 = 3670 (moved three decimal points)

0.005 × 1000,000 = 5000 (moved three decimal points)

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Step-by-step explanation:

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3 years ago

Evaluate the limit

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We are given with a limit and we need to find it's value so let's start !!!!

${\quad \qquad \blacktriangleright \blacktriangleright \displaystyle \sf \lim_{x\to 4}\dfrac{\sqrt{x}-\sqrt{3\sqrt{x}-2}}{x^{2}-16}}$

But , before starting , let's recall an identity which is the main key to answer this question

${\boxed{\bf{a^{2}-b^{2}=(a+b)(a-b)}}}$

Consider The limit ;

${:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{\sqrt{x}-\sqrt{3\sqrt{x}-2}}{x^{2}-16}}$

Now as directly putting the limit will lead to indeterminate form 0/0. So , Rationalizing the numerator i.e multiplying both numerator and denominator by the conjugate of numerator 

${:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{\sqrt{x}-\sqrt{3\sqrt{x}-2}}{x^{2}-16}\times \dfrac{\sqrt{x}+\sqrt{3\sqrt{x}-2}}{\sqrt{x}+\sqrt{3\sqrt{x}-2}}}$

${:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-\sqrt{3\sqrt{x}-2})(\sqrt{x}+\sqrt{3\sqrt{x}-2})}{(x^{2}-4^{2})(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}$

Using the above algebraic identity ;

${:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x})^{2}-(\sqrt{3\sqrt{x}-2})^{2}}{(x-4)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}$

${:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{x-(3\sqrt{x}-2)}{(x-4)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}$

${:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{x-3\sqrt{x}+2}{\{(\sqrt{x})^{2}-2^{2}\}(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}$

${:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{x-3\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}$

Now , here we need to eliminate (√x-2) from the denominator somehow , or the limit will again be indeterminate ,so if you think carefully as I thought after seeing the question i.e what if we add 4 and subtract 4 in numerator ? So let's try !

${:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{x-3\sqrt{x}-2+4-4}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}$

${:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(x-4)+2+4-3\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}$

Now , using the same above identity ;

${:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-2)(\sqrt{x}+2)+6-3\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}$

${:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-2)(\sqrt{x}+2)+3(2-\sqrt{x})}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}$

Now , take minus sign common in numerator from 2nd term , so that we can take (√x-2) common from both terms

${:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-2)(\sqrt{x}+2)-3(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}$

Now , take (√x-2) common in numerator ;

${:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-2)\{(\sqrt{x}+2)-3\}}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}$

Cancelling the radical that makes our limit again and again indeterminate ;

${:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{\cancel{(\sqrt{x}-2)}\{(\sqrt{x}+2)-3\}}{\cancel{(\sqrt{x}-2)}(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}$

${:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}+2-3)}{(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}$

${:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-1)}{(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}$

Now , putting the limit ;

${:\implies \quad \sf \dfrac{\sqrt{4}-1}{(\sqrt{4}+2)(4+4)(\sqrt{4}+\sqrt{3\sqrt{4}-2})}}$

${:\implies \quad \sf \dfrac{2-1}{(2+2)(4+4)(2+\sqrt{3\times 2-2})}}$

${:\implies \quad \sf \dfrac{1}{(4)(8)(2+\sqrt{6-2})}}$

${:\implies \quad \sf \dfrac{1}{(4)(8)(2+\sqrt{4})}}$

${:\implies \quad \sf \dfrac{1}{(4)(8)(2+2)}}$

${:\implies \quad \sf \dfrac{1}{(4)(8)(4)}}$

${:\implies \quad \sf \dfrac{1}{128}}$

${:\implies \quad \bf \therefore \underline{\underline{\displaystyle \bf \lim_{x\to 4}\dfrac{\sqrt{x}-\sqrt{3\sqrt{x}-2}}{x^{2}-16}=\dfrac{1}{128}}}}$

3 0

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ki77a [65]

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8 0

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