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Katen [24]
3 years ago
5

What value of x makes this equation true?

Mathematics
1 answer:
Arlecino [84]3 years ago
5 0

Answer:

x = 4

Step-by-step explanation:

Given

13 - 4x = 1 - x ( add x to both sides )

13 - 3x = 1 ( subtract 13 from both sides )

- 3x = - 12 ( divide both sides by - 3 )

x = 4

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3 years ago
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

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2 years ago
How can I create an independent and dependent probability word problem that is about gumball? (also, the answer has to be writte
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A jar of jelly beans contains 50 red gumballs , 45 yellow gumballs, and 30 green gumballs. You reach into the jar and randomly select a jelly bean, then select another without putting the first jelly bean back. What is the probability that you draw two red jelly beans? This is Dependent because you didnt put the other jelly bean in thus changing the total nmber of jelly beans.

A jar of jelly beans contains 50 red gumballs<span> , 45 yellow gumballs, and 30 green gumballs. You reach into the jar and randomly select a jelly bean, then select another while replacing  the first jelly bean back. What is the probability that you draw two red jelly beans? This is Independent because you put the other jelly bean in thus keeping the total number of jelly beans.</span>

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Which of the following properties was used in the expression shown here?
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Distributive property
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Help meh pls- 5x(20 divided by 4)to the power of 2
vagabundo [1.1K]

Answer:

125

Step-by-step explanation:

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