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zmey
2 days ago
6

Which function has a simplified base of 4RootIndex 3 StartRoot 4 EndRoot? f(x) = 2(RootIndex 3 StartRoot 16 EndRoot) Superscript

x f(x) = 2(RootIndex 3 StartRoot 64 EndRoot) Superscript x f(x) = 4(RootIndex 3 StartRoot 16 EndRoot) Superscript 2 x f(x) = 4(RootIndex 3 StartRoot 64 EndRoot) Superscript 2 x

Mathematics
2 answers:
tester [8.8K]2 days ago
9 0

Response:

  f(x)=4\sqrt[3]{16}^{2x}

Detailed explanation:

You're likely in search of a function with a base that can be simplified to...

  4\sqrt[3]{4}\approx 6.3496

The functions you seem to be considering appear to be...

  f(x)=2\sqrt[3]{16}^x\approx 2\cdot2.5198^x\\\\f(x)=2\sqrt[3]{64}^x=2\cdot 4^x\\\\f(x)=4\sqrt[3]{16}^{2x}\approx 4\cdot 6.3496^x\ \leftarrow\text{ this one}\\\\f(x)=4\sqrt[3]{64}^{2x}=4\cdot 16^x

It looks like the third option is the one that fits your requirements.

Leona [9.2K]2 days ago
5 0

Response: refer to the image

Detailed explanation:

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tester [8842]

Answer:

Step-by-step explanation:

Denote the rectangular prism's volume as VR and the volume of the right triangular prism as VT. Given that the rectangular prism's volume exceeds that of the triangular prism by 32 cubic feet, we establish that VR = 32 + VT.

Calculating VR: VR = length * width * height = 6*x*3.

This simplifies to VR = 18x ft³.

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Now, simplify the equation:

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This results in 4x = 32.

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x = 8.

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Answer:

Explanation:

Hello!

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