# Thread: Help with a few quick questions (Only 14 year olds work so easy)

1. ## Help with a few quick questions (Only 14 year olds work so easy)

1) For the two numbers "-2 and 8", it is impossibl to find the geometric mean. Is this true or false? And explain your answer

2)Write the missing numbers in this fraction sum:
"1 OVER 3" PLUS "8 OVER X" X=?

3) The area of face of a cube is 9xSQUARED.
Write and expression for the total surface rea of the cube. Write your answer as simply as possible

Also, write an expression for the volume of the cube. Also as simply as possible.

4) The fraction "1 OVER 9" is half of the fraction ________

THNK YOU!

2. 1) For the two numbers "-2 and 8", it is impossibl to find the geometric mean. Is this true or false? And explain your answer

The geometric mean only applies to positive numbers in order to avoid taking the root of a negative product, thus yielding an imaginary number. In your example, you would have $\sqrt{-16}$. Not possible.

2)Write the missing numbers in this fraction sum:
"1 OVER 3" PLUS "8 OVER X" X=?

$\frac{1}{3}+\frac{8}{x}= ??$ Something missing here, I think.

3) The area of face of a cube is 9xSQUARED.
Write and expression for the total surface rea of the cube. Write your answer as simply as possible

Total surface would be the sum of the areas of all 6 faces. Therefore, $SA=6(9x^2)=54x^2$

Also, write an expression for the volume of the cube. Also as simply as possible.

Volume of a cube is expressed as $V=s^3$

Each edge would have to be $3x$, so

$V=(3x)^3=27x^3$

4) The fraction "1 OVER 9" is half of the fraction ________

$\frac{1}{9}=\frac{1}{2}x$

$\frac{1}{9}=\frac{x}{2}$

$9x=2$

$x=\frac{2}{9}$

..

3. Wow ultimate thanks man (THANKED)

In number two, the whole question is as follows:

Write the missing numbers in these fraction sums

A) "1 OVER 4" PLUS "x OVER 8" = 1

I got x as 6

B) "1 OVER 3" PLUS "8 OVER X" = 1

Does that help?

4. Originally Posted by FB FTW
Wow ultimate thanks man (THANKED)

In number two, the whole question is as follows:

Write the missing numbers in these fraction sums

A) "1 OVER 4" PLUS "x OVER 8" = 1

I got x as 6

B) "1 OVER 3" PLUS "8 OVER X" = 1

Does that help?
$\frac{1}{4}+\frac{x}{8}=1$

Multiply everything by LCD of 8

$2+x=8$

$x=6$

You were correct!!
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$\frac{1}{3}+\frac{8}{x}$

Multiply everything by LCD of 3x

$x+24=3x$

$24=2x$

$x=12$