Questions: Imagine you derive the following expression by analyzing the physics of a particular system: a=g sin theta - muk g cos theta, where g=9.80 meter per second squared. Simplify the expression for a by pulling out the common factor. a=(9.80 meter / second squared) sin theta - muk(9.80 meter / second squared) cos theta a=g sin theta - muk g cos theta (no simplification should be performed on the expression in this situation) a=g(sin theta - muk cos theta)

Imagine you derive the following expression by analyzing the physics of a particular system: a=g sin theta - muk g cos theta, where g=9.80 meter per second squared. Simplify the expression for a by pulling out the common factor. a=(9.80 meter / second squared) sin theta - muk(9.80 meter / second squared) cos theta a=g sin theta - muk g cos theta (no simplification should be performed on the expression in this situation) a=g(sin theta - muk cos theta)
Transcript text: Imagine you derive the following expression by analyzing the physics of a particular system: $a=g \sin \theta-\mu_{k} g \cos \theta$, where $g=9.80$ meter $^{\text {Simplity }}$ second ${ }^{2}$. Simplify the expression for $a$ by pulling out the common factor. $a=\left(9.80\right.$ meter $/$ second $\left.{ }^{2}\right) \sin \theta-\mu_{k}\left(9.80\right.$ meter $/$ second $\left.{ }^{2}\right) \cos \theta$ $a=g \sin \theta-\mu_{k} g \cos \theta$ (no simplification should be performed on the expression in this situation) $a=g\left(\sin \theta-\mu_{k} \cos \theta\right)$
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Solution

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Simplifying the expression for \( a \) by pulling out the common factor.

Identify the common factor

In the expression \( a = g \sin \theta - \mu_{k} g \cos \theta \), the common factor is \( g \).

Factor out the common factor

By factoring out \( g \), the expression becomes \( a = g(\sin \theta - \mu_{k} \cos \theta) \).

\(\boxed{a = g(\sin \theta - \mu_{k} \cos \theta)}\)

The simplified expression for \( a \) by pulling out the common factor is \(\boxed{a = g(\sin \theta - \mu_{k} \cos \theta)}\).

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