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4 years ago

The voltage, V, produced by an ac generator at time, t, in seconds, is V(t)=220sin(120π(t−1360)). Find the period of V(t).

Mathematics

1 answer:

patriot [66]4 years ago

5 0

ANSWER

The period is

$\frac{1}{60}$

EXPLANATION

The equation that models the voltage, V, produced in an AC generator at time, t, in seconds is

$V(t)=220 \sin(120\pi(t-1360))$

The period of this function is given by:

$\frac{2 \pi}{B}$

Substitute the value of B to get;

$= \frac{2\pi}{120\pi}$

Simplify

$= \frac{1}{60}$

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1052 toothpicks can be grouped into 4 groups of third power of 6 ( $6^{3}$ ), 5 groups of second power of 6 ( $6^{2}$ ), 1 group of first power of 6 ( $6^{1}$ ) and 2 groups of zeroth power of 6 ( $6^{0}$ ).

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Given: 1052 toothpicks

To do: The objective is to group the toothpicks in powers of 6 and to write the number 1052 as a base 6 number

First we note that, $6^{0}=1,6^{1}=6,6^{2}=36,6^{3}=216,6^{4}=1296$

This implies that $6^{4}$ exceeds 1052 and thus the highest power of 6 that the toothpicks can be grouped into is 3.

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So, after grouping the toothpicks into 4 groups of third power of 6, there are 188 toothpicks remaining.

Now, $6^{2}=36$ and $36\times 5=180, 36\times 6=216$ . This implies that $36\times 6$ exceeds 188 and thus there can be at most 5 groups of $6^{2}$ .

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$188-5\times6^{2}$

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So, after grouping the remaining toothpicks into 5 groups of second power of 6, there are 8 toothpicks remaining.

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So, after grouping the remaining toothpicks into 1 group of first power of 6, there are 2 toothpicks remaining.

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This concludes the grouping.

Thus, it was obtained that 1052 toothpicks can be grouped into 4 groups of third power of 6 ( $6^{3}$ ), 5 groups of second power of 6 ( $6^{2}$ ), 1 group of first power of 6 ( $6^{1}$ ) and 2 groups of zeroth power of 6 ( $6^{0}$ ).

Then,

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