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ivolga24 [154]
3 years ago
10

Fourth roots of 81(cos320°+isin320°)

Mathematics
1 answer:
sattari [20]3 years ago
7 0
I am the Glob-glo-gab-galab
<span>The shwabble-dabble-wabble-gabble flibba blabba blab
I'm full of shwibbly glib-a-kind</span>
I am the yeast of thoughts and minds
<span>Shwabble dabble glibble glabble schribble shwap glab
Dibble dabble shribble shrabble glibbi-glap shwap
Shwabble dabble glibble glabble shwibble shwap-dap
Dibble dabble shribble shrabble glibbi-shwap glab</span>

[Bridge]
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Simply delicious
Ooooohm, ha haa ha ha

[Chorus]
I am the Glob-glo-gab-galab
<span>The shwabble-dabble-wabble-gabble flibba blabba blab
I'm full of shwibbly glib-a-kind</span>
I am the yeast of thoughts and minds
<span>Shwabble dabble glibble glabble schribble shwap glab
Dibble dabble shribble shrabble glibbi-glap shwap
Shwabble dabble glibble glabble shwibble shwap-dap
Dibble dabble shribble shrabble glibbi-shwap glab</span>
<span />
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The interest on a $6,000, 6%, 60-day note receivable is
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The circle shown below has a diameter of 18 centimeters. What is the area of the shaded sector?
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If you had 1052 toothpicks and were asked to group them in powers of 6, how many groups of each power of 6 would you have? Put t
sukhopar [10]

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}).

The number 1052, written as a base 6 number is 4512

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.

Now, 6^{3}=216 and 216\times 5=1080, 216\times 4=864. This implies that 216\times 5 exceeds 1052 and thus there can be at most 4 groups of 6^{3}.

Then,

1052-4\times6^{3}

1052-4\times216

1052-864

188

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}.

Then,

188-5\times6^{2}

188-5\times36

188-180

8

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

Now, 6^{1}=6 and 6\times 1=6, 6\times 2=12. This implies that 6\times 2 exceeds 8 and thus there can be at most 1 group of 6^{1}.

Then,

8-1\times6^{1}

8-1\times6

8-6

2

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

Now, 6^{0}=1 and 1\times 2=2. This implies that the remaining toothpicks can be exactly grouped into 2 groups of zeroth power of 6.

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,

1052=4\times6^{3}+5\times6^{2}+1\times6^{1}+2\times6^{0}

So, the number 1052, written as a base 6 number is 4512.

Learn more about change of base of numbers here:

brainly.com/question/14291917

6 0
2 years ago
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