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Oksi-84 [34.3K]
3 years ago
5

Write , or = to make the statement true. 9.008 ___ 9.08

Mathematics
1 answer:
laila [671]3 years ago
7 0

Answer:

9.008 < 9.08

Step-by-step explanation:

9.08 is greater because you compare the number 9 and 9 are the same, 0 and 0 still the same, but then 0 and 8 you know eight is greater. was this helpful enough

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Answer:

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Step-by-step explanation:

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2/5(a+2b)^2 (if a=3 and b=-4)
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The answer is 10. Hope this helps.

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aleksklad [387]
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Stella [2.4K]

Answer:  -\frac{\sqrt{2a}}{8a}

=======================================================

Explanation:

The (x-a) in the denominator causes a problem if we tried to simply directly substitute in x = a. This is because we get a division by zero error.

The trick often used for problems like this is to rationalize the numerator as shown in the steps below.

\displaystyle \lim_{x\to a} \frac{\sqrt{3a-x}-\sqrt{x+a}}{4(x-a)}\\\\\\\lim_{x\to a} \frac{(\sqrt{3a-x}-\sqrt{x+a})(\sqrt{3a-x}+\sqrt{x+a})}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{(\sqrt{3a-x})^2-(\sqrt{x+a})^2}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{3a-x-(x+a)}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{3a-x-x-a}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\

\displaystyle \lim_{x\to a} \frac{2a-2x}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{-2(-a+x)}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{-2(x-a)}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{-2}{4(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\

At this point, the (x-a) in the denominator has been canceled out. We can now plug in x = a to see what happens

\displaystyle L = \lim_{x\to a} \frac{-2}{4(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\L = \frac{-2}{4(\sqrt{3a-a}+\sqrt{a+a})}\\\\\\L = \frac{-2}{4(\sqrt{2a}+\sqrt{2a})}\\\\\\L = \frac{-2}{4(2\sqrt{2a})}\\\\\\L = \frac{-2}{8\sqrt{2a}}\\\\\\L = \frac{-1}{4\sqrt{2a}}\\\\\\L = \frac{-1*\sqrt{2a}}{4\sqrt{2a}*\sqrt{2a}}\\\\\\L = \frac{-\sqrt{2a}}{4\sqrt{2a*2a}}\\\\\\L = \frac{-\sqrt{2a}}{4\sqrt{(2a)^2}}\\\\\\L = \frac{-\sqrt{2a}}{4*2a}\\\\\\L = -\frac{\sqrt{2a}}{8a}\\\\\\

There's not much else to say from here since we don't know the value of 'a'. So we can stop here.

Therefore,

\displaystyle \lim_{x\to a} \frac{\sqrt{3a-x}-\sqrt{x+a}}{4(x-a)} = -\frac{\sqrt{2a}}{8a}\\\\\\

3 0
3 years ago
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