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How do you add #sqrt32+9sqrt2-sqrt18#?

Answer 1

Julian Avila

#sqrt32+9sqrt2-sqrt18=10sqrt2#

#sqrt32+9sqrt2-sqrt18#

= #sqrt(ul(2xx2xx2xx2)xx2)+9sqrt2-sqrt(ul(3xx3)xx2)#

= #2xx2xxsqrt2+9sqrt2-3sqrt2#

= #4sqrt2+9sqrt2-3sqrt2#

= #(4+9-3)sqrt2#

= #10sqrt2#

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Answer 2

Nova Allen

To add ( \sqrt{32} + 9\sqrt{2} - \sqrt{18} ), first, we need to simplify each term:

( \sqrt{32} ) can be simplified as ( \sqrt{16 \times 2} = \sqrt{16} \times \sqrt{2} = 4\sqrt{2} ).
( \sqrt{18} ) can be simplified as ( \sqrt{9 \times 2} = \sqrt{9} \times \sqrt{2} = 3\sqrt{2} ).

Now, substitute the simplified terms back into the expression:

( 4\sqrt{2} + 9\sqrt{2} - 3\sqrt{2} )

Combine like terms:

( (4 + 9 - 3)\sqrt{2} )

( 10\sqrt{2} )

Therefore, ( \sqrt{32} + 9\sqrt{2} - \sqrt{18} = 10\sqrt{2} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.