5 answers
To simplify the expression \( \sqrt{25} \cdot \sqrt{49} \), we can first calculate each square root separately.
\[ \sqrt{25} = 5 \] \[ \sqrt{49} = 7 \]
Now, we multiply these results together:
\[ \sqrt{25} \cdot \sqrt{49} = 5 \cdot 7 = 35 \]
Thus, the product of the expression in simplest radical form is
\[ \boxed{35} \]
To simplify the expression \( \sqrt{25} \cdot \sqrt{49} \), we can use the property of square roots that states \( \sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b} \).
First, we compute:
\[ \sqrt{25} = 5 \] \[ \sqrt{49} = 7 \]
Thus, we combine these:
\[ \sqrt{25} \cdot \sqrt{49} = 5 \cdot 7 = 35 \]
Alternatively, applying the property of square roots:
\[ \sqrt{25} \cdot \sqrt{49} = \sqrt{25 \cdot 49} \]
Calculating \( 25 \cdot 49 \):
\[ 25 \cdot 49 = 1225 \]
Then we find the square root:
\[ \sqrt{1225} = 35 \]
So, both methods confirm that the product of the expression in simplest radical form is
\[ \boxed{35} \]
To simplify the expression \( \frac{8\sqrt{27}}{2\sqrt{45}} \), we start by simplifying both the numerator and the denominator.
Simplify the numerator: \[ 8\sqrt{27} = 8\sqrt{9 \cdot 3} = 8\sqrt{9}\sqrt{3} = 8 \cdot 3 \cdot \sqrt{3} = 24\sqrt{3} \]
Simplify the denominator: \[ 2\sqrt{45} = 2\sqrt{9 \cdot 5} = 2\sqrt{9}\sqrt{5} = 2 \cdot 3 \cdot \sqrt{5} = 6\sqrt{5} \]
Rewrite the expression using the simplified forms: \[ \frac{8\sqrt{27}}{2\sqrt{45}} = \frac{24\sqrt{3}}{6\sqrt{5}} \]
Simplify the fraction: \[ \frac{24}{6} = 4 \] Thus, \[ \frac{24\sqrt{3}}{6\sqrt{5}} = 4\cdot\frac{\sqrt{3}}{\sqrt{5}} \]
Rationalize the denominator: To rationalize, we multiply the numerator and denominator by \(\sqrt{5}\): \[ 4 \cdot \frac{\sqrt{3}}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = 4 \cdot \frac{\sqrt{15}}{5} = \frac{4\sqrt{15}}{5} \]
Thus, the quotient of the expression in simplest radical form, with the denominator rationalized, is
\[ \boxed{\frac{4\sqrt{15}}{5}} \]
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