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NeTakaya
3 years ago
10

We have a bilateral triangle with base length 2 and sidelengths x where x is increasing at a rate

Mathematics
1 answer:
BaLLatris [955]3 years ago
7 0

Answer:

\frac{5\sqrt{3} }{6}

Step-by-step explanation:

Given :

base length = 2

side lengths = x

dx/dt = 5 unit/min

Determine d∅ / dt  at x = 2

d∅ / dt  at x = 2  = \frac{5\sqrt{3} }{6}

attached below is a detailed solution

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The equivalent expression to the givan expression is

\sqrt[4]{324m^{12}}\times\sqrt[3]{64k^9}=12\sqrt[4]{4}m^3k^3

Step-by-step explanation:

Given expression is 4th root of 324m^12 * the cubed root of 64k^9

The given expression can be written as

\sqrt[4]{324m^{12}}\times\sqrt[3]{64k^9}

To find the equivalent expression to the given expression :

\sqrt[4]{324m^{12}}\times\sqrt[3]{64k^9}

=\sqrt[4]{81\times 4m^{12}}\times\sqrt[3]{16\times 4k^9}

=\sqrt[4]{3\times 3\times 3\times 3\times 4m^{12}}\times\sqrt[3]{4\times 4\times 4k^9}

=\sqrt[4]{3\times 3\times 3\times 3\times 4m^{12}}\times\sqrt[3]{4\times 4\times 4k^9}

=(3\times \sqrt[4]{m^{12}})\times (4\times \sqrt[3]{k^9})

=(3\times \sqrt[4]{4m^{12}})\times (4\times \sqrt[3]{k^9})

=(3\times \sqrt[4]{4}\times (m^{12})^{\frac{1}{4}})\times (4\times {(k^9)^{\frac{1}{3}})

=(3\sqrt[4]{4}\times m^{\frac{12}{4}})\times (4\times k^{\frac{9}{3}})

=(3\sqrt[4]{4}\times m^3)\times (4\times k^3)

=3\sqrt[4]{4}m^3.4k^3

=12\sqrt[4]{4}m^3k^3

\sqrt[4]{324m^{12}}\times\sqrt[3]{64k^9}=12\sqrt[4]{4}m^3k^3

Therefore the equivalent expression to the given expression is

\sqrt[4]{324m^{12}}\times\sqrt[3]{64k^9}=12\sqrt[4]{4}m^3k^3

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