(\sqrt32 + \sqrt128) \times \sqrt2 = ?

NIACL Model Paper 7 with solutions for free online practice

Question : 37 of 100

Marks: +1, -0

(√32 + √128) × √2 = ?

Solution:

(4)

? = (√32+√128)×√2

= (4√2+8√2)×√2

= 12√2 × √2 =24

Go to Question:

Prev Question

Next Question