New user? Sign up

Existing user? Sign in

If we have \(xy = 7, yz = 3, xz = 5\), then the numerical expression \(3x^2 + 5y^2 + 7z^2\) is equal to:

Note by Mahla Salarmohammadi 1 year, 11 months ago

Easy Math Editor

*italics*

_italics_

**bold**

__bold__

- bulleted- list

1. numbered2. list

paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)

> This is a quote

This is a quote

# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

2 \times 3

2^{34}

a_{i-1}

\frac{2}{3}

\sqrt{2}

\sum_{i=1}^3

\sin \theta

\boxed{123}

Sort by:

\(xy=7\) .....\((1)\) \(yz=3\) .....\((2)\) \(xz=5\) .....\((3)\)

From \((1)\) and \((2)\).

\(x=\frac{7}{y}=\frac{5}{z}\) \(y=\frac{7z}{5}\)

Putting the value of \(y\) in \((3)\).

\(\Rightarrow y^2=\boxed{\frac{21}{5}}\) From \((1)\) and \((2)\). \(y=\frac{3}{z}=\frac{7}{x}\) \(y=3x\) and \(x=\frac{7z}{3}\)

Putting the value of \(y\) in \((1)\).

\(\Rightarrow x^2=\boxed{\frac{7}{3}}\)

Putting the value of \(x\) in \((3)\).

\(\Rightarrow z^2=\boxed{\frac{15}{7}}\)

Now,

\(3×\frac{7}{3}+5×\frac{21}{5}+7×\frac{15}{7}\) \(7+21+15=\boxed{43}.\)

Log in to reply

I sorry. I only know the answer

Ok.Then tell me where I made the mistake in my solution.

Your answer is wrong!! Right answer:71😊

Then,where I made the mistake.Please show your solution.

Thanks you for your helping.are you sure?

Yes. I am sure.

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewest\(xy=7\) .....\((1)\)

\(yz=3\) .....\((2)\)

\(xz=5\) .....\((3)\)

From \((1)\) and \((2)\).

\(x=\frac{7}{y}=\frac{5}{z}\)

\(y=\frac{7z}{5}\)

Putting the value of \(y\) in \((3)\).

\(\Rightarrow y^2=\boxed{\frac{21}{5}}\)

From \((1)\) and \((2)\).

\(y=\frac{3}{z}=\frac{7}{x}\)

\(y=3x\) and \(x=\frac{7z}{3}\)

Putting the value of \(y\) in \((1)\).

\(\Rightarrow x^2=\boxed{\frac{7}{3}}\)

Putting the value of \(x\) in \((3)\).

\(\Rightarrow z^2=\boxed{\frac{15}{7}}\)

Now,

\(3×\frac{7}{3}+5×\frac{21}{5}+7×\frac{15}{7}\)

\(7+21+15=\boxed{43}.\)

Log in to reply

I sorry. I only know the answer

Log in to reply

Ok.Then tell me where I made the mistake in my solution.

Log in to reply

Your answer is wrong!! Right answer:71😊

Log in to reply

Then,where I made the mistake.Please show your solution.

Log in to reply

Thanks you for your helping.are you sure?

Log in to reply

Yes. I am sure.

Log in to reply