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ella [17]
3 years ago
15

Hi I really need help on this problem. Thank you!

Mathematics
1 answer:
Olin [163]3 years ago
3 0

Answer:

The circumference of a circle with radius 13.8 is 86.71

Step-by-step explanation:

Circumference of a cicle in terms of radius:

Circumference = 2·π·r = 2·3.14·13.8 = 86.71

In terms of diameter:

Circumference = π·d = 3.14·27.6 = 86.66

In terms of area:

Circumference C = √4·π·A = √4·π·598.3 = 86.71

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Nikitich [7]
The answer to this question is x=4.
4 0
3 years ago
Can someone help please (highlighted in blue)
Annette [7]

\implies {\blue {\boxed {\boxed {\purple {\sf {B)\:10}}}}}}

\large\mathfrak{{\pmb{\underline{\red{Step-by-step\:explanation}}{\red{:}}}}}

4 + 3 \: ( \: 10 -  {2}^{3}  \: )

➺\: 4 + 3 \: ( \: 10 - 2 \times 2 \times 2 \: )

➺\: 4 + 3 \: ( \: 10 - 8 \: )

➺\: 4 + 3 \: ( \: 2 \: )

➺\: 4 + 6

➺\: 10

<h3><u>Note</u>:-</h3>

\sf\pink{PEMDAS\: rule.}

P = Parentheses

E = Exponents

M = Multiplication

D = Division

A = Addition

S = Subtraction

\large\mathfrak{{\pmb{\underline{\orange{Mystique35 }}{\orange{❦}}}}}

6 0
3 years ago
Read 2 more answers
The bar graph shows the number of school days Jalen had homework and did not have homework during the first six months of
Oksanka [162]

Given:

<u>In July:</u>

Number of days he has done the homework = 6

Number of days he has not done the homework = 4

<u>In August:</u>

Number of days he has done the homework = 16

Number of days he has not done the homework = 7

<u>In September:</u>

Number of days he has done the homework = 14

Number of days he has not done the homework = 4

<u>In October:</u>

Number of days he has done the homework = 14

Number of days he has not done the homework = 8

<u>In November:</u>

Number of days he has done the homework = 16

Number of days he has not done the homework = 4

<u>In December:</u>

Number of days he has done the homework = 5

Number of days he has not done the homework = 10

To find the months in which the difference between the number of days with homework and with no homework greater than 6.

Now,

<u>In July,</u>

The difference between the number of days with homework and with no homework = 6-4 = 2

<u>In August,</u>

The difference between the number of days with homework and with no homework = 16-7 = 9

<u>In September,</u>

The difference between the number of days with homework and with no homework = 14-4 = 10

<u>In October,</u>

The difference between the number of days with homework and with no homework = 14-8 = 6

<u>In November,</u>

The difference between the number of days with homework and with no homework = 16-4 = 12

<u>In December,</u>

The difference between the number of days with homework and with no homework = 10-5 = 5

Hence,

In 3 months the difference between the number of days with homework and with no homework greater than 6 and these are: August, September, November.

4 0
3 years ago
How to solve 14x + 2 = 6 (x - 0)
Sergeu [11.5K]

Answer:

Has not solution.

Step-by-step explanation:

14x÷2 = 6(x - 0)

14x÷2 = 6*x + 6*-0

14x÷2 = 6x - 0

14x = 2*6x

14x ≠ 12x

then:

This equation has not solution.

8 0
3 years ago
The list below shows the Highlight in inches of the player on a man college basketball team
boyakko [2]

We have the data set:

{79, 83, 82, 78, 79, 77, 80, 73, 82, 72}

This data set has 10 items.

To complete the data summary is helpful to sort the data:

{72, 73, 77, 78, 79, 79, 80, 82, 82, 83}

The minimum value is 72.

<em>The quartile divides the data set in 4 parts. </em>

<em>The first quartile is the value for which 25% of the data is below.</em>

<em>The second quartile is the value for which 50% of the data is below. The second quartile is equivalent to the median.</em>

<em>The third quartile is the value for which 75% of the data is below.</em>

In this data set, as its size is a even number, the quartile fall between two positions, so we will calculate the average of this numbers.

The first quartile will fall in the position 0.25*10=2.5. Then we can calculate the 1st quartile as the average between the 2nd and 3rd number of the data set.

This values are 73 (2nd position) and 77 (3rd position), so the average is:

Q_1=\frac{73+77}{2}=75

We can repeat this with the other quartiles:

\begin{gathered} Q_2=M=\frac{79+79}{2}=79 \\ Q_3=\frac{82+82}{2}=82 \end{gathered}

Then:

The first quartile is 75.

The median (second quartile) is 79.

The third quartile is 82.

The maximum value is 83.

7 0
1 year ago
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