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2 years ago

What's the answer to this?

Mathematics

2 answers:

Ymorist [56]2 years ago

4 0

Answer: 9

Step-by-step explanation:

Natalija [7]2 years ago

3 0

Answer:

Step-by-step explanation:

We can see that 6 is 2 more than 3, and so the difference is -3^2, 9.

We can check this by seeing that -3^6 is -729 and -3^4 is -81.

We have

-729 / -81 = 9.

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2 years ago

Because of their connection with secant lines, tangents, and instantaneous rates, limits of the form ModifyingBelow lim With h

Gre4nikov [31]

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$\dfrac{1}{2\sqrt{x}}$

Step-by-step explanation:

$f(x) = \sqrt{x} = x^{\frac{1}{2}}$

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This can be rewritten as

$[x(1+\dfrac{h}{x})]^{\frac{1}{2}}$

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Setting $x=\dfrac{h}{x}$ and $n=\frac{1}{2}$ ,

$(1+\dfrac{h}{x})^{\frac{1}{2}}=1+(\dfrac{h}{x})(\dfrac{1}{2})+\dfrac{\frac{1}{2}(1-\frac{1}{2})}{2!}(\dfrac{h}{x})^2+\tldots$

$=1+\dfrac{h}{2x}-\dfrac{h^2}{8x^2}+\ldots$

Multiplying by $x^{\frac{1}{2}}$ ,

$x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}=x^{\frac{1}{2}}+\dfrac{h}{2x^{\frac{1}{2}}}-\dfrac{h^2}{8x^{\frac{3}{2}}}+\ldots$

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$\dfrac{x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}-x^{\frac{1}{2}}}{h}=\dfrac{1}{2x^{\frac{1}{2}}}-\dfrac{h}{8x^{\frac{3}{2}}}+\ldots$

The limit of this as $h\to 0$ is

$\lim_{h\to0} \dfrac{f(x+h)-f(x)}{h}=\dfrac{1}{2x^{\frac{1}{2}}}=\dfrac{1}{2\sqrt{x}}$ (since all the other terms involve $h$ and vanish to 0.)

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