Vinculum

From my knowledge of geometry, I don't think a rectangle has 8 vertices......

Can you please rephrase your question

60.5

$\sqrt{x\sqrt{x\sqrt{x}}}=\sqrt{x^{1+\frac{1}{2}+\frac{1}{4}}}$

$\sqrt{x\sqrt{x\sqrt{x}}}=\sqrt{x^{\frac{7}{4}}}$

The ANSWER IS: $\sqrt{x^{\frac{7}{4}}}$

I am actually bad at this kind of stuff so anyone... PLEASE CORRECT ME IF I AM WRONG:

I got $522 for the first option, $226.11 for the second option and $375.24 for the third option.

The best option would be the second option.

Yeah man, this kind of stuff should be talked about privately......

I'm in 9th grade

55,000/300 = 183.33333333333

Die Antwort ist: 183.33333333333

I'm pretty sure that 4 is the answer, but....hold up

Wouldn't that make it a square though??

Can anyone please confirm?

From the information given above, we can say that the width of the rectangle is $4$

The area of a rectangle is $length \cdot width$

That is, $4 \cdot length = 16$

That gives us length = $16/4 = 4$

From that, we can tell that the value of $x$ must be $4$.