Question

simplify.
$5\sqrt{147}$
$\sqrt{3}$
$\sqrt{49}$
$\sqrt{35}$

Explanation:

Step1: Factor 147 into prime factors

We know that \(147 = 49\times3\), and \(49 = 7^2\). So we can rewrite \(\sqrt{147}\) as \(\sqrt{49\times3}\).
\[
\sqrt{147}=\sqrt{49\times3}
\]

Step2: Use the property of square roots \(\sqrt{ab}=\sqrt{a}\times\sqrt{b}\) (where \(a = 49\), \(b = 3\))

According to the property \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\) (\(a\geq0,b\geq0\)), we have \(\sqrt{49\times3}=\sqrt{49}\times\sqrt{3}\). Since \(\sqrt{49} = 7\), then \(\sqrt{49\times3}=7\sqrt{3}\).
\[
\sqrt{147}=7\sqrt{3}
\]

Step3: Multiply by 5

Now we have the original expression \(5\sqrt{147}\), substitute \(\sqrt{147}=7\sqrt{3}\) into it, we get \(5\times7\sqrt{3}\).
\[
5\sqrt{147}=5\times7\sqrt{3}
\]

Step4: Calculate \(5\times7\)

Calculate \(5\times7 = 35\), so \(5\times7\sqrt{3}=35\sqrt{3}\).
\[
5\times7\sqrt{3}=35\sqrt{3}
\]

Answer:

\(35\sqrt{3}\)