What is the $3 \sqrt{125}$ $3sqrt125$ minus $2 \sqrt{80}$ $2sqrt80$ plus $\sqrt{405}$ $sqrt405$ ?

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Answer 1 · 2018-04-14T22:04:55

$16 \sqrt{5}$ $16sqrt5$

$3 \sqrt{125} - 2 \sqrt{80} + \sqrt{405}$ $3sqrt125-2sqrt80+sqrt405$

$3 \sqrt{25 \cdot 5} - 2 \sqrt{16 \cdot 5} + \sqrt{81 \cdot 5}$ $3sqrt(25*5)-2sqrt(16*5)+sqrt(81*5)$

$3 \cdot 5 \sqrt{5} - 2 \cdot 4 \sqrt{5} + 9 \sqrt{5}$ $3*5sqrt5-2*4sqrt5+9sqrt5$

$15 \sqrt{5} - 8 \sqrt{5} + 9 \sqrt{5}$ $15sqrt5-8sqrt5+9sqrt5$

$7 \sqrt{5} + 9 \sqrt{5}$ $7sqrt5+9sqrt5$

$16 \sqrt{5}$ $16sqrt5$

What is the 3√1253sqrt125 minus 2√802sqrt80 plus √405sqrt405?