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AleksAgata [21]
4 years ago
11

Can someone help me with this question, I got it wrong.

Mathematics
1 answer:
lukranit [14]4 years ago
8 0

Answer:

8.29

Step-by-step explanation:

degrees of sector / by 360 * by pi* r^2

You might be interested in
Step 2 - Fill in the missing number: A vertical line and horizontal line combine to make a L shape. There is one row of entries
aivan3 [116]

The synthetic division's representation of the dividend is 2x3 + 10x2 + x + 5.

Given that

An L shape is created when two lines intersect vertically and horizontally.

The shape has entries in two rows.

Entries in row 1 are 2, 10, 1, and 5.

Blank, -10, and 0 are the entries in row 2.

A simplified method of dividing a polynomial with another polynomial equation of degree one is known as synthetic division.

On the exterior, to the left of the form, is entry number 5.

The entry stands for the divisor's zero.

If the variable is x, then this entry to the variable is;

2x³+10x²+x+5

The dividend is thus represented by synthetic division as

2x³+10x²+x+5

The Question is incomplete And complete question is given below!!

What dividend is represented by the synthetic division below? A vertical line and horizontal line combine to make a L shape. There are two rows of entries within the shape. Row 1 has entries 2, 10, 1, 5. Row 2 has entries blank, negative 10, 0, negative 5. Entry negative 5 is on the outside to the left of the shape, and a third row of entries is outside and below the shape. Row 3 has entries 2, 0, 1, 0. Negative 10 x squared minus 5 2 x cubed 10 x squared x 5 2 x squared 1 2 x cubed x.

Learn more about Synthetic division here

brainly.com/question/27979968

#SPJ1

5 0
2 years ago
Is 4x positive or negative?
Nikolay [14]

Answer:

if x is positive, the answer is positive. If x is negative, the answer is negative.

7 0
3 years ago
Given the function f(x) = x2 and k = 2, which of the following represents a horizontal shift?
melamori03 [73]
A horizontal shift would be in the form <span><span>f(x±k)</span></span>
6 0
3 years ago
Read 2 more answers
Layla played sports for 3 3/4 hours. Punish time playing sports was 3/8 as long. About how many hours did puno play sports
Nostrana [21]
I would say 13 3/4 but dont take me for my word
4 0
3 years ago
Mystery Boxes: Breakout Rooms
ollegr [7]

Answer:

\begin{array}{ccccccccccccccc}{1} & {3} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {57} & {58} & {61} \\ \end{array}

Step-by-step explanation:

Given

\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {[ \ ]} \\ \end{array}

Required

Fill in the box

From the question, the range is:

Range = 60

Range is calculated as:

Range =  Highest - Least

From the box, we have:

Least = 1

So:

60 = Highest  - 1

Highest = 60 +1

Highest = 61

The box, becomes:

\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}

From the question:

IQR = 20 --- interquartile range

This is calculated as:

IQR = Q_3 - Q_1

Q_3 is the median of the upper half while Q_1 is the median of the lower half.

So, we need to split the given boxes into two equal halves (7 each)

<u>Lower half:</u>

\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } \\ \end{array}

<u>Upper half</u>

<u></u>\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}<u></u>

The quartile is calculated by calculating the median for each of the above halves is calculated as:

Median = \frac{N + 1}{2}th

Where N = 7

So, we have:

Median = \frac{7 + 1}{2}th = \frac{8}{2}th = 4th

So,

Q_3 = 4th item of the upper halves

Q_1= 4th item of the lower halves

From the upper halves

<u></u>\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}<u></u>

<u></u>

We have:

Q_3 = 32

Q_1 can not be determined from the lower halves because the 4th item is missing.

So, we make use of:

IQR = Q_3 - Q_1

Where Q_3 = 32 and IQR = 20

So:

20 = 32 - Q_1

Q_1 = 32 - 20

Q_1 = 12

So, the lower half becomes:

<u>Lower half:</u>

\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {12 } & {15} & {18}& {[ \ ] } \\ \end{array}

From this, the updated values of the box is:

\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {12} & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}

From the question, the median is:

Median = 22 and N = 14

To calculate the median, we make use of:

Median = \frac{N + 1}{2}th

Median = \frac{14 + 1}{2}th

Median = \frac{15}{2}th

Median = 7.5th

This means that, the median is the average of the 7th and 8th items.

The 7th and 8th items are blanks.

However, from the question; the mode is:

Mode = 18

Since the values of the box are in increasing order and the average of 18 and 18 do not equal 22 (i.e. the median), then the 7th item is:

7th = 18

The 8th item is calculated as thus:

Median = \frac{1}{2}(7th + 8th)

22= \frac{1}{2}(18 + 8th)

Multiply through by 2

44 = 18 + 8th

8th = 44 - 18

8th = 26

The updated values of the box is:

\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}

From the question.

Mean = 26

Mean is calculated as:

Mean = \frac{\sum x}{n}

So, we have:

26= \frac{1 + 2nd + 4 + 12 + 15 + 18 + 18 + 26 + 29 + 30 + 32 + 12th + 58 + 61}{14}

Collect like terms

26= \frac{ 2nd + 12th+1 + 4 + 12 + 15 + 18 + 18 + 26 + 29 + 30 + 32 + 58 + 61}{14}

26= \frac{ 2nd + 12th+304}{14}

Multiply through by 14

14 * 26= 2nd + 12th+304

364= 2nd + 12th+304

This gives:

2nd + 12th = 364 - 304

2nd + 12th = 60

From the updated box,

\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}

We know that:

<em>The 2nd value can only be either 2 or 3</em>

<em>The 12th value can take any of the range 33 to 57</em>

Of these values, the only possible values of 2nd and 12th that give a sum of 60 are:

2nd = 3

12th = 57

i.e.

2nd + 12th = 60

3 + 57 = 60

So, the complete box is:

\begin{array}{ccccccccccccccc}{1} & {3} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {57} & {58} & {61} \\ \end{array}

6 0
3 years ago
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