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Shalnov [3]
2 years ago
8

Find the area of the regular 18-gon with radius 11 mm

Mathematics
1 answer:
Anna35 [415]2 years ago
3 0

Answer:

36

Step-by-step explanation:

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How many students can sit around a cluster of 7 square table? The tables in a classroom have square tops. Four students can comf
Sveta_85 [38]

Answer:

16 students can sit around a cluster of 7 square table.

Step-by-step explanation:

Consider the provided information.

We need to find how many students can sit around a cluster of 7 square table.

The tables in a classroom have square tops.

Four students can comfortably sit at each table with ample working space.

If we put the tables together in cluster it will look as shown in figure.

From the pattern we can observe that:

Number of square table in each cluster   Total number of students

                     1                                                                 4

                     2                                                                6

                     3                                                                8

                     4                                                                10

                     5                                                                12

                     6                                                                14

                     7                                                                16

Hence, 16 students can sit around a cluster of 7 square table.

8 0
3 years ago
Read 2 more answers
5x - x = -12 what is x???
aleksley [76]
X would equal -3. 5x-3 would equal -15, and 2 negatives would equal a positive. So, -15+3 would equal -12.
3 0
3 years ago
Read 2 more answers
write in scientific  notation 1. 0.0000364 2. 0.00751 3. 0.10005 4. 1,094 5. 0.00000099 6. 0.04101 7. 10,500 8. 8,9000
solmaris [256]
1.\\0\underbrace{.00003}_{5\to}54=3.54\times10^{-5}\\2.\\0\underbrace{.007}_{3\to}51=7.51\times10^{-3}\\3.\\0\underbrace{.1}_{1\to}0005=1.0005\times10^{-1}\\4.\\1\underbrace{,094}_{\leftarrow3}=1.094\times10^3\\5.\\0\underbrace{.0000009}_{7\to}9=9.9\times10^{-7}
6.\\0\underbrace{.04}_{2\to}101=4.101\times10^{-2}\\7.\\1\underbrace{0,500}_{\leftarrow4}=1.05\times10^4\\8.\\8\underbrace{9000}_{\leftarrow4}=8.9\times10^4
3 0
2 years ago
Mrs. brook lacks 7 years from being 5 times as old as her son. 6 years from now she will lack 3 years from being 3 times as old
fomenos
Is this a true of false question if so yes
5 0
3 years ago
Boris choos three diffrent numbers.the sum of the three numbers is 36.One of trh numbers is a cube number.the other two number a
inna [77]

Answer:

4,5,27

Problem:

Boris chose three different numbers.

The sum of the three numbers is 36.

One of the numbers is a perfect cube.

The other two numbers are factors of 20.

Step-by-step explanation:

Let's pretend those numbers are:

a,b, \text{ and } c.

We are given the sum is 36: a+b+c=36.

One of our numbers is a perfect cube. a=n^3 where n is an integer.

The other two numbers are factors of 20. bk=20 and ci=20 where a,c,i, \text{ and } k \text{ are integers}.

n^3+\frac{20}{k}+\frac{20}{i}=36

From here I would just try to find numbers that satisfy the conditions using trial and error.

3^3+\frac{20}{2}+\frac{20}{2}

27+10+10

47

3^3+\frac{20}{4}+\frac{20}{5}

27+5+4

36

So I have found a triple that works:

27,5,4

The numbers in ascending order is:

4,5,27

4 0
3 years ago
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