Waste less time on Facebook — follow Brilliant.
×

Algebra

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

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> 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"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

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

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,

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

Abhay Kumar - 1 year, 11 months ago

Log in to reply

I sorry. I only know the answer

Mahla Salarmohammadi - 1 year, 11 months ago

Log in to reply

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

Abhay Kumar - 1 year, 11 months ago

Log in to reply

Your answer is wrong!! Right answer:71😊

Mahla Salarmohammadi - 1 year, 11 months ago

Log in to reply

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

Abhay Kumar - 1 year, 11 months ago

Log in to reply

Thanks you for your helping.are you sure?

Mahla Salarmohammadi - 1 year, 11 months ago

Log in to reply

Yes. I am sure.

Abhay Kumar - 1 year, 11 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...