Question

What is the sign of


\[\dfrac{x^{59}}{2.3x}\cdot\dfrac{4}5\] when
\[x < 0\]?
Choose 1 answer:
Choose 1 answer:
(Choice A) Positive
A
Positive
(Choice B) Negative
B
Negative
(Choice C) Zero
C
Zero

Answer 1

To determine the sign of the expression

\[
\dfrac{x^{59}}{2.3x}\cdot\dfrac{4}{5}
\]

when \(x < 0\), we can simplify the expression.

First, let's simplify the fraction \(\dfrac{x^{59}}{2.3x}\):

\[
\dfrac{x^{59}}{2.3x} = \dfrac{x^{59}}{x} \cdot \dfrac{1}{2.3} = \dfrac{x^{58}}{2.3}
\]

Now, we include the factor \(\dfrac{4}{5}\):

\[
\dfrac{x^{58}}{2.3} \cdot \dfrac{4}{5} = \dfrac{4x^{58}}{11.5}
\]

Now, let's analyze \(x^{58}\):

Since \(x < 0\), the power \(58\) (which is even) will result in \(x^{58}\) being positive. Therefore, \(x^{58} > 0\).

Now we can analyze the entire expression:

1. \(4\) is positive.
2. \(11.5\) is positive.
3. Thus, \(\dfrac{4x^{58}}{11.5}\) is of the form \(\text{(positive)} \cdot \text{(positive)}\).

This results in the expression being positive.

So, the sign of

\[
\dfrac{x^{59}}{2.3x}\cdot\dfrac{4}{5}
\]

when \(x < 0\) is:

\(\text{(Choice A) Positive}\).