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3 years ago

Can someone please explain this.

Mathematics

1 answer:

Vitek1552 [10]3 years ago

5 0

Answer:

The correct option is option A

Step-by-step explanation:

We have the following expression:

sqrt(18x^3) - sqrt(9x^3) + 3sqrt(x^3) - sqrt(2x^3)

We know that sqrt(a*b) = sqrt(a)sqrt(b)

Applying this, we have:

sqrt(18)sqrt(x^3) - 3sqrt(x^3) + 3sqrt(x^3) - sqrt(2)sqrt(x^3)

sqrt(x^3)[ sqrt(18) - sqrt(2)]

sqrt(x^3)[ 3sqrt(2)- sqrt(2)]

sqrt(x^3)[2sqrt(2)]

Then we now that:

sqrt(x^3)[2sqrt(2)] = 2x*sqrt(2x)

The correct option is option A

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Answer:

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$=\int^1_0[\int^3_0([\frac{6cos x^2 \times \sqrt z}{5\times \frac{1}{2}}]^9_{3y})dy]dx$

$=\int^1_0[\int^3_0([\frac{12cos x^2 \times( \sqrt 9-\sqrt{3y})}{5}])dy]dx$

$=\int^1_0[\int^3_0([\frac{12cos x^2 \times( 3-\sqrt{3y})}{5}])dy]dx$

$=\int^1_0[\frac{12cos x^2 \times( 3y-\frac{\sqrt{3}y^\frac{3}{2}}{\frac{3}{2}})}{5}]^3_0dx$

$=\int^1_0[\frac{12cos x^2 \times( 3.3-\frac{2\sqrt{3}.3^\frac{3}{2}}{3})}{5}]^3_0dx$

$=\int^1_0[\frac{12cos x^2 \times( 9-6)}{5}]dx$

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