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Vlada [557]
3 years ago
7

Which of the following is a step in constructing a circle circumscribed about a triangle

Mathematics
1 answer:
sweet-ann [11.9K]3 years ago
7 0

Answer:

Construct the perpendicular bisectors of each side

Choice A is correct

Step-by-step explanation:

Circumscribe means to draw on the outside of a geometric figure just touching the corner points but never crossing.

The steps in constructing a circle circumscribed about a triangle;

Construct the perpendicular bisector of one side

Construct the perpendicular bisector of another side

Where the lines cross is the center of the Circumscribed circle .

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Violet Flowergarden- What is 2 x 2?
babunello [35]

Answer:

4

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
A square is put together by two triangles. The hypotenuse is 11 cm and width of 8 cm. find the height of triangle?
Aleks [24]
You can use the pythagorean theorem. Make sure that you put the measurements in the right places though. Since 11 is the measure of the hypotenuse, 11 has to be c. When you plug in the information that you have, it should look like this:
8^2 + b^2 = 11^2

Square 8 and 11. 
64 + b^2 = 121

Subtract 64 from 121
121 - 64 = 57

The new equation is:
b^2 = 57

Now take the square root of both sides to get b by itself. 
b = about 7.5

Answer: The height of the triangle is 7.5

***** Note: I rounded the answer ********************************************
4 0
3 years ago
Which term completes the product so that it is the difference of squares?
FromTheMoon [43]

Answer:

the answer is  3

Step-by-step explanation:

(-5x-3)(-5x+3)=(5x)^2 - (3)^2

A difference of squares is the difference of two squared terms.

We have

a^2 - b^2

We can factorize the difference of squared terms like this:

a^2-b^2 = (a+b)(a-b)

We have (-5x-3)(-5x+3)

Lets prove it:

(-5x-3)(-5x+3) = (-5x*-5x)+(-5x*3)+(-3*-5x)+(-3*3)

(-5x-3)(-5x+3) = 25x^2+(-15x)+(15x)+(-9)

(-5x-3)(-5x+3) = 25x^2-15x+15x-9

(-5x-3)(-5x+3) = 25x^2-9

(-5x-3)(-5x+3) = (5x)^2 - (3)^2

6 0
3 years ago
Read 2 more answers
Two soccer teams play 8 games in their season. The number of goals each team scored per game is listed below: Team X: 11, 3, 0,
Gre4nikov [31]

Answer:

C. Team Y’s scores have a lower mean value.

Step-by-step explanation:

We are given that Two soccer teams play 8 games in their season. The number of goals each team scored per game is listed below:

Team X: 11, 3, 0, 0, 2, 0, 6, 4

Team Y: 4, 2, 0, 3, 2, 1, 6, 4

Firstly, we will calculate the mean, median, range and inter-quartile range for Team X;

Mean of Team X data is given by the following formula;

        Mean, \bar X =  \frac{\sum X}{n}

                       =  \frac{11+ 3+ 0+ 0+ 2+ 0+ 6+ 4}{8}  =  \frac{26}{8}  = 3.25

So, the mean of Team X's scores is 3.25.

Now, for calculating the median; we have to arrange the data in ascending order and then observe that the number of observations (n) in the data is even or odd.

Team X: 0, 0, 0, 2, 3, 4, 6, 11

  • If n is odd, then the formula for calculating median is given by;

                         Median  =  (\frac{n+1}{2} )^{th} \text{ obs.}

  • If n is even, then the formula for calculating median is given by;

                         Median  =  \frac{(\frac{n}{2})^{th} \text{ obs.} +(\frac{n}{2}+1)^{th} \text{ obs.}  }{2}

Here, the number of observations is even, i.e. n = 8.

So, Median =  \frac{(\frac{n}{2})^{th} \text{ obs.} +(\frac{n}{2}+1)^{th} \text{ obs.}  }{2}

                   =  \frac{(\frac{8}{2})^{th} \text{ obs.} +(\frac{8}{2}+1)^{th} \text{ obs.}  }{2}

                   =  \frac{(4)^{th} \text{ obs.} +(5)^{th} \text{ obs.}  }{2}

                   =  \frac{2+3}{2}  = 2.5

So, the median of Team X's score is 2.5.

Now, the range is calculated as the difference between the highest and the lowest value in our data.

               Range = Highest value - Lowest value

                           = 11 - 0 = 11

So, the range of Team X's score is 11.

Now, the inter-quartile range of the data is given by;

        Inter-quartile range = Q_3-Q_1

Q_1=(\frac{n+1}{4} )^{th} \text{ obs.}

     =  (\frac{8+1}{4} )^{th} \text{ obs.}

     =  (2.25 )^{th} \text{ obs.}

Q_1 = 2^{nd} \text{ obs.} + 0.25[ 3^{rd} \text{ obs.} -2^{nd} \text{ obs.} ]

     =  0 + 0.25[0 - 0] = 0

Q_3=3(\frac{n+1}{4} )^{th} \text{ obs.}

     =  3(\frac{8+1}{4} )^{th} \text{ obs.}

     =  (6.75 )^{th} \text{ obs.}

Q_3 = 6^{th} \text{ obs.} + 0.75[ 7^{th} \text{ obs.} -6^{th} \text{ obs.} ]

     =  4 + 0.75[6 - 4] = 5.5

So, the inter-quartile range of Team X's score is (5.5 - 0) = 5.5.

<u>Now, we will calculate the mean, median, range and inter-quartile range for Team Y;</u>

Mean of Team Y data is given by the following formula;

        Mean, \bar Y =  \frac{\sum Y}{n}

                       =  \frac{4+ 2+ 0+ 3+ 2+ 1+ 6+ 4}{8}  =  \frac{22}{8}  = 2.75

So, the mean of Team Y's scores is 2.75.

Now, for calculating the median; we have to arrange the data in ascending order and then observe that the number of observations (n) in the data is even or odd.

Team Y: 0, 1, 2, 2, 3, 4, 4, 6

  • If n is odd, then the formula for calculating median is given by;

                         Median  =  (\frac{n+1}{2} )^{th} \text{ obs.}

  • If n is even, then the formula for calculating median is given by;

                         Median  =  \frac{(\frac{n}{2})^{th} \text{ obs.} +(\frac{n}{2}+1)^{th} \text{ obs.}  }{2}

Here, the number of observations is even, i.e. n = 8.

So, Median =  \frac{(\frac{n}{2})^{th} \text{ obs.} +(\frac{n}{2}+1)^{th} \text{ obs.}  }{2}

                   =  \frac{(\frac{8}{2})^{th} \text{ obs.} +(\frac{8}{2}+1)^{th} \text{ obs.}  }{2}

                   =  \frac{(4)^{th} \text{ obs.} +(5)^{th} \text{ obs.}  }{2}

                   =  \frac{2+3}{2}  = 2.5

So, the median of Team Y's score is 2.5.

Now, the range is calculated as the difference between the highest and the lowest value in our data.

               Range = Highest value - Lowest value

                           = 6 - 0 = 6

So, the range of Team Y's score is 6.

Now, the inter-quartile range of the data is given by;

        Inter-quartile range = Q_3-Q_1

Q_1=(\frac{n+1}{4} )^{th} \text{ obs.}

     =  (\frac{8+1}{4} )^{th} \text{ obs.}

     =  (2.25 )^{th} \text{ obs.}

Q_1 = 2^{nd} \text{ obs.} + 0.25[ 3^{rd} \text{ obs.} -2^{nd} \text{ obs.} ]

     =  1 + 0.25[2 - 1] = 1.25

Q_3=3(\frac{n+1}{4} )^{th} \text{ obs.}

     =  3(\frac{8+1}{4} )^{th} \text{ obs.}

     =  (6.75 )^{th} \text{ obs.}

Q_3 = 6^{th} \text{ obs.} + 0.75[ 7^{th} \text{ obs.} -6^{th} \text{ obs.} ]

     =  4 + 0.75[4 - 4] = 4

So, the inter-quartile range of Team Y's score is (4 - 1.25) = 2.75.

Hence, the correct statement is:

C. Team Y’s scores have a lower mean value.

4 0
3 years ago
In a parallelogram, consecutive angles are
goldenfox [79]

Answer:

Consecutive angles will always some to 180 degrees, that makes consecutive angles in a parallelogram supplementary

5 0
3 years ago
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