Solve:

$$\frac{1}{\sqrt[3]{4x}}$$

$$\frac{1}{\sqrt[3]{4x}}$$

Bruce Sherman
2022-10-01
Answered

Solve:

$$\frac{1}{\sqrt[3]{4x}}$$

$$\frac{1}{\sqrt[3]{4x}}$$

You can still ask an expert for help

Giancarlo Phelps

Answered 2022-10-02
Author has **10** answers

Solution:

$$\frac{1}{\sqrt[3]{4x}}=\frac{1}{(4x{)}^{1/3}}\phantom{\rule{0ex}{0ex}}\frac{1}{\sqrt[3]{4x}}=\frac{1}{(4x{)}^{1/3}}\phantom{\rule{0ex}{0ex}}=(4x{)}^{-\frac{1}{3}}$$

$$\frac{1}{\sqrt[3]{4x}}=\frac{1}{(4x{)}^{1/3}}\phantom{\rule{0ex}{0ex}}\frac{1}{\sqrt[3]{4x}}=\frac{1}{(4x{)}^{1/3}}\phantom{\rule{0ex}{0ex}}=(4x{)}^{-\frac{1}{3}}$$

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What is the fraction $\frac{4}{9}$ simplified?

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What are the ways to calculate the difference between fractions where the denominators are different?

For example,

$\frac{4}{9}\frac{27}{54}$

For example,

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How to rearrange: $16=\frac{1}{n}25+\frac{n-1}{n}218.75$

Can anyone here help me out to rearrange the following formula and solve for $n$?

Can anyone here help me out to rearrange the following formula and solve for $n$?

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How to find the value of a and b from this limit problem with or without L'Hopital's formula?

Consider the limit:

$\underset{x\to 4}{lim}\frac{{x}^{2}+ax+b}{x-4}=14$

Question: How can I find the values of a and b?

Attempt:

My first thought is, we need to use L'Hopital's rule to make sure that the denominator isn't zero:

Applying L'Hopital's rule, and we get:

$\underset{x\to 4}{lim}\frac{2x+a}{1}=14$

Then, we can substitute the limit of x to the equation such that:

$2(4)+a=8+a=14$

and we get that the value of a is 6.

But, how can I find the value of b? It seems that after applying the L'Hopital's formula the value of b disappears.

Also, is there a way to solve this problem without L'Hopital's rule?

Thanks

Consider the limit:

$\underset{x\to 4}{lim}\frac{{x}^{2}+ax+b}{x-4}=14$

Question: How can I find the values of a and b?

Attempt:

My first thought is, we need to use L'Hopital's rule to make sure that the denominator isn't zero:

Applying L'Hopital's rule, and we get:

$\underset{x\to 4}{lim}\frac{2x+a}{1}=14$

Then, we can substitute the limit of x to the equation such that:

$2(4)+a=8+a=14$

and we get that the value of a is 6.

But, how can I find the value of b? It seems that after applying the L'Hopital's formula the value of b disappears.

Also, is there a way to solve this problem without L'Hopital's rule?

Thanks

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Rearranging exponential functions

Using two different strategies, I've derived an equation for a particular function $f(\varphi )$. That equation is

$f(\varphi )=\frac{1}{1-{e}^{-T\gamma}}(1-{e}^{-T\gamma \varphi}).$

However, the paper whose result I'm trying to replicate is telling me that the function is

$f(\varphi )=(1-{e}^{-\gamma})(1-{e}^{-\gamma \varphi}).$

Is there a way to rearrange the result I got so that it matches the result in the paper?

EDIT:

What's confusing me about this is that the function needs to be such that $f(0)=0$ and $f(1)=1$. This is true of the equation I derived. But for the equation in the paper, $f(1)=1$ only if $\gamma =\mathrm{ln}(1/2)$, which is not the case in the specific example given later in the paper. So something weird is afoot (either on my part or on the paper's).

Using two different strategies, I've derived an equation for a particular function $f(\varphi )$. That equation is

$f(\varphi )=\frac{1}{1-{e}^{-T\gamma}}(1-{e}^{-T\gamma \varphi}).$

However, the paper whose result I'm trying to replicate is telling me that the function is

$f(\varphi )=(1-{e}^{-\gamma})(1-{e}^{-\gamma \varphi}).$

Is there a way to rearrange the result I got so that it matches the result in the paper?

EDIT:

What's confusing me about this is that the function needs to be such that $f(0)=0$ and $f(1)=1$. This is true of the equation I derived. But for the equation in the paper, $f(1)=1$ only if $\gamma =\mathrm{ln}(1/2)$, which is not the case in the specific example given later in the paper. So something weird is afoot (either on my part or on the paper's).

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If the chance of an event was $1/128$ and increased by 20%, what is the new chance?

So I have something that has a 1/128 chance of occurring, let's say. Suddenly, the chances of that thing happening are increased by 20%. How is that fraction written? Would you multiply 1/128 by 6/5 (yielding 3/320), or would you take 128, multiply it by 4/5, and then invert it (1/102)?

This isn't for homework, I'm merely curious.

Thank you

So I have something that has a 1/128 chance of occurring, let's say. Suddenly, the chances of that thing happening are increased by 20%. How is that fraction written? Would you multiply 1/128 by 6/5 (yielding 3/320), or would you take 128, multiply it by 4/5, and then invert it (1/102)?

This isn't for homework, I'm merely curious.

Thank you

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find the rational number halfway between the numbers 7/8 and 2/3