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katovenus [111]
3 years 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:
PIT_PIT [208]3 years ago
9 0

Answer:

  f(x)=4\sqrt[3]{16}^{2x}

Step-by-step explanation:

We believe you're wanting to find a function with an equivalent base of ...

  4\sqrt[3]{4}\approx 6.3496

The functions you're looking at seem 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

The third choice seems to be the one you're looking for.

diamong [38]3 years ago
5 0

Answer:look at the pic

Step-by-step explanation:

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Answer:

B . Yes

Step-by-step explanation:

Recall: a tangent of a circle is perpendicular to the radius of a circle, forming a right angle at the point of tangency.

Therefore, if segment ST is tangent to circle P, it means that m<T = 90°, and ∆PST is a right triangle.

To know if ∆PST is a right triangle, the side lengths should satisfy the Pythagorean triple rule, which is:

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Step-by-step explanation:

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