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erma4kov [3.2K]
3 years ago
13

Can y’all help me on Thai thank you

Mathematics
1 answer:
elena-14-01-66 [18.8K]3 years ago
7 0

Answer:

=8400

Step-by-step explanation:

(1860+240)(4)

=8400

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Write four related facts. use pictures to show how the number sentences are related 1st grade
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Any answer choices?
5 0
3 years ago
Ill give a brainliest if you can answer all three :))
Pavel [41]

Answer:

4. A

8. D

9. B

Step-by-step explanation:

4. started with 10 points: +10

lost 20 points: -20

won 45 points: +45

10-20+45

8. 11-(-8)

=19

9. 100-(-50)-2

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6 0
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Solve each equation<br> l x+11 l=8<br><br> l 3x-2 l=16
kobusy [5.1K]
|x + 11| = 8
 x + 11 = 8
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|3x - 2| = 16
3x + 2 = 16
<u>      -2     -2</u>
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      3      3
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3 0
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

6 0
3 years ago
Write an equation of the graph
serious [3.7K]

Answer:

the answer should be -1 to 3

Step-by-step explanation:

or it could take another form -3> 4

3 0
3 years ago
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