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nika2105 [10]
3 years ago
5

Please give me the correct answer

Mathematics
1 answer:
vladimir1956 [14]3 years ago
6 0

Answer:

10 centimeters

Step-by-step explanation:

formula for volume of a cylinder = πr² · h

1. Set up the equation

(3.14)(r²)(14) = 4,396

2. Simplify

(43.96)(r²) = 4,396

3. Solve

r² = 100

√r = √100

r = 10

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Please answer this correctly
PIT_PIT [208]

Which variable is dependent on another variable's value?

-The number of squares Rosa cuts the brownies into because it is dependent on the number of people attending the potluck.

Therefore, the independent variable is p, the number of people attending the potluck.

Hope this helps! :)

7 0
3 years ago
G+5>12 solution please i don’t know how to do this
Musya8 [376]

Answer:  g > 7

Graph has an open circle at 7 on the number line, shading to the right

================================================

Explanation:

Think of it like saying "I have a number, and I add on 5. The result is something larger than 12". You can guess and check your way to the answer, but the quickest way is to subtract 5 from both sides.

We subtract to undo the addition happening to the 'g'.

g+5 > 12

g+5-5 > 12-5

g > 7

So the number is larger than 7. For instance, if g = 8, then,

g+5 > 12

8+5 > 12

13 > 12

This is a true statement.

-------------------------

If you need to graph the solution, then you'll have an open circle at 7 on the number line. The open circle says to the reader "don't include this value as part of the solution set". Shade to the right of the open circle to describe all values larger than 7.

In summary, the graph has an open circle at 7 and shading to the right.

3 0
3 years ago
Write the number 0.21212121…. as a fraction.
vekshin1

100x = 21.212121.....

100x - x = 21.212121... – 0.212121...

99x = 21. ( divide both side by 90)

x = 21/99

x = 21:3/99:3

x = 7/33

<h3><u>PLEASE</u><u> MARK</u><u> ME</u><u> BRAINLIEST</u><u>.</u></h3>
6 0
3 years ago
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Delvig [45]

Answer: B&E

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4 0
4 years ago
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sasho [114]

Answer:

1/5 pound

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6 0
3 years ago
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