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amm1812
3 years ago
11

Please help me with 21 and 22! Thank you.

Mathematics
1 answer:
Artyom0805 [142]3 years ago
7 0
#21

To find the area of rectangle, you multiple length by width, so that would be (\ \sqrt{2x}\ )(\ \sqrt{6x}\ ) = \sqrt{2x\cdot6x} = \sqrt{12x^2} = x\sqrt{4\cdot3} = \boxed{2x\sqrt3}

Since we are talking about area \ length, x need to be positive so we don't need to worry about absolute value now.

#22

The area of triangle is \dfrac12bh, so that would be
\dfrac12\sqrt{18y}\cdot\sqrt{2y^4}\\\ \\=\dfrac{y^2}2\sqrt{18y}\cdot\sqrt{2}\\\ \\=\dfrac{y^2}2\sqrt{18y\cdot2}\\\ \\=\dfrac{y^2}2\sqrt{36y}\\\ \\=\dfrac{y^2}2\sqrt{4\cdot9\cdot y}\\\ \\=\boxed{3y^2\sqrt y}\text{ or }\boxed{3\sqrt{y^5}}

Hope this helps.
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