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abruzzese [7]
2 years ago
9

A city has a population of 220,000 people. Suppose that each year the population grows by 6.25%. What will the population be aft

er 11 years? Use the calculator provided and round your answer to the nearest whole number.
Mathematics
1 answer:
Murrr4er [49]2 years ago
6 0

Your answer will be 137500

Im not 100% sure, im 90% sure

Hope this helps :)

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Rita has a rope 5 m long. She cuts it into 2 pieces of length 2 m 30 cm and 1 m 65 cm.
Diano4ka-milaya [45]

Answer:

3m 95cm

Step-by-step explanation:

its... simple math? i added 2 and 1; and 30 and 65?

6 0
3 years ago
I NEED HELP ON NUMBER 6 PLEASE A AND B, ITS
olya-2409 [2.1K]
P-111
A-90???

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6 0
3 years ago
Solve the inequality, <br> x + 6.56 &gt; 25.2
pogonyaev

Answer: x>18.64

Step-by-step explanation:

x+6.56-6.56>25.2-6.56

x>18.64

7 0
3 years ago
Type the integer that makes the following addition sentence true:<br> = 6<br> 39 +
quester [9]

-33

Hope this helps. :)))))

8 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
2 years ago
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