Does anyone happen to know a nice way to show that $(a+b{)}^{n}\le {a}^{n}+{b}^{n}$, where $a,b\ge 0$ and $n\in (0,1]$? I figured integrating might help, but I've been unable to pull my argument full circle. Any suggestions are appreciated :)

Almintas2l
2022-07-17
Answered

How to show $(a+b{)}^{n}\le {a}^{n}+{b}^{n}$, where $a,b\ge 0$ and $n\in (0,1]$?

Does anyone happen to know a nice way to show that $(a+b{)}^{n}\le {a}^{n}+{b}^{n}$, where $a,b\ge 0$ and $n\in (0,1]$? I figured integrating might help, but I've been unable to pull my argument full circle. Any suggestions are appreciated :)

Does anyone happen to know a nice way to show that $(a+b{)}^{n}\le {a}^{n}+{b}^{n}$, where $a,b\ge 0$ and $n\in (0,1]$? I figured integrating might help, but I've been unable to pull my argument full circle. Any suggestions are appreciated :)

You can still ask an expert for help

asked 2022-05-26

If $\frac{a}{b}=\frac{b}{c}=\frac{c}{d}$, prove that $\frac{a}{d}=\sqrt{\frac{{a}^{5}+{b}^{2}{c}^{2}+{a}^{3}{c}^{2}}{{b}^{4}c+{d}^{4}+{b}^{2}c{d}^{2}}}$

What I've done so far;

$\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\phantom{\rule{0ex}{0ex}}a=bk,b=ck,c=dk\phantom{\rule{0ex}{0ex}}a=c{k}^{2},b=d{k}^{2}\phantom{\rule{0ex}{0ex}}a=d{k}^{3}$

I tried substituting above values in the right hand side of the equation to get $\frac{a}{d}$ but couldn't.

What I've done so far;

$\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\phantom{\rule{0ex}{0ex}}a=bk,b=ck,c=dk\phantom{\rule{0ex}{0ex}}a=c{k}^{2},b=d{k}^{2}\phantom{\rule{0ex}{0ex}}a=d{k}^{3}$

I tried substituting above values in the right hand side of the equation to get $\frac{a}{d}$ but couldn't.

asked 2022-07-07

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

asked 2022-06-25

Algebraic Problem Fractions

Given the following fraction,

$\frac{n}{{2}^{n-1}}(-2{)}^{n-1}$

Is it correct that it simplifies to the following fraction,

$(-1{)}^{n-1}n$

I am in doubt because I have checked on three different places and they all gave different answer.

Given the following fraction,

$\frac{n}{{2}^{n-1}}(-2{)}^{n-1}$

Is it correct that it simplifies to the following fraction,

$(-1{)}^{n-1}n$

I am in doubt because I have checked on three different places and they all gave different answer.

asked 2021-08-11

For each of the sets below, determine whether {2} is an element of that set.

a)

b)

c)

d)

e)

f)

asked 2022-07-04

Can fractions be written with a one under them and can problems be solved this way

How would you write $\frac{a}{0.1}$?

Are either of these ways acceptable?

$\frac{\frac{a}{1}}{\frac{\frac{1}{10}}{1}}=\frac{a}{1}\times \frac{1}{\frac{1}{10}}$

or

$\frac{\frac{a}{1}}{\frac{1}{10}}=\frac{a}{1}\times \frac{10}{1}$

How would you write $\frac{a}{0.1}$?

Are either of these ways acceptable?

$\frac{\frac{a}{1}}{\frac{\frac{1}{10}}{1}}=\frac{a}{1}\times \frac{1}{\frac{1}{10}}$

or

$\frac{\frac{a}{1}}{\frac{1}{10}}=\frac{a}{1}\times \frac{10}{1}$

asked 2022-05-13

What must be added to this mixed fraction to produce this mixed fraction

This may seem like a simple question which is fit for a Google search; however, I'm unable to find an answer.

What must be added to $6\frac{3}{5}$ to produce $3\frac{5}{6}$

I hope I'll finally understand. Step-by-step solution is much appreciated.

This may seem like a simple question which is fit for a Google search; however, I'm unable to find an answer.

What must be added to $6\frac{3}{5}$ to produce $3\frac{5}{6}$

I hope I'll finally understand. Step-by-step solution is much appreciated.

asked 2021-11-16

Simplify the complex fraction.

$\frac{\frac{20}{7x}}{\frac{10}{21x}}=?$