Question

If root 3 = 1.732, then the value of root27-3root75+5root 48 + 2 root 108

Answer 1

Step-by-step explanation:

Solution :

$\: \: \: \sqrt{27} - 3 \sqrt{75} + 5 \sqrt{48} + 2 \sqrt{108}$

$= > \sqrt{9 \times 3} -3 \sqrt{25 \times 3} + 5 \sqrt{16 \times 3} + 2 \sqrt{36 \times 3}$

$= > 3 \sqrt{3} - 3 \times 5 \sqrt{3} + 5 \times 4 \sqrt{3} + 2 \times 6 \sqrt{3}$

$= > 3 \sqrt{3} - 15 \sqrt{3} + 20 \sqrt{3} + 12 \sqrt{3}$

$= > (3 - 15 + 20 + 12) \sqrt{3}$

$= > 20 \times \sqrt{3}$

$= > 20 \times 1.732$

$= > 34.64$

Hence :

The value is 34.64

Answer 2

Answer:

Heya user,

Here is the answer you were looking for:

$\begin{lgathered}\sqrt{27} - 3 \sqrt{75} + 5 \sqrt{48} + 2 \sqrt{108} \\\end{lgathered} </p><p>27$

$\begin{lgathered}\sqrt{3 \times 3 \times 3} - 3 \sqrt{5 \times 5 \times 3} + 4 \sqrt{2 \times 2 \times 2 \times 2 \times 3} + 2 \sqrt{2 \times 2 \times 3 \times 3 \times 3} \\ \\ = \sqrt{ {3}^{2} \times 3} - 3 \sqrt{ {5}^{2} \times 3 } + 4 \sqrt{ {2}^{2} \times {2}^{2} \times 3} + 2 \sqrt{ {2}^{2} \times {3}^{2} \times 3 } \\ \\ = 3 \sqrt{3} - 3 \times 5 \sqrt{3} + 4 \times 2 \times 2\sqrt{3} + 2 \times 2 \times 3 \sqrt{3} \\ \\ = 3 \sqrt{3} - 15 \sqrt{3} + 16 \sqrt{3} + 12 \sqrt{3} \\ \\ = 31 \sqrt{3} - 15 \sqrt{3} \\ \\ = 16 \sqrt{3} \\ \\ = 16 \times 1.732 \\ \\ = 27.712\end{lgathered} </p><p>3×3×3$

Hope this helps!!