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Hatshy [7]
3 years ago
15

HELP!!! Find the lengths of the diagonals of this trapezoid.

Mathematics
1 answer:
Alina [70]3 years ago
5 0

Answer:

√[ (a + b)^2 + c^2 ].

Step-by-step explanation:

They are both equal in length because we have an isosceles trapezoid here. We see form the drawing it is symmetrical about the y axis.

The diagonal = the line joining points (b, c) and (-a, 0)

= √[ (b - -a)^2 + (c - 0)^2 ]

= √[ (a + b)^2 + c^2].

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Find the area of the polygon​
Evgesh-ka [11]

Answer:

784m^2

Step-by-step explanation:

is a rectangle with sides of 40m and 20m, in the upper right corner a right triangle has been removed with the legs of: 40 - 32 = 8m and 20 - 16 = 4m, we find the area of ​​the rectangle (b * h).

40 * 20 = 800m ^ 2, then we find the area of ​​the right triangle

1/2 b * h: 1/2 8 * 4 = 16 m ^ 2.

we remove the area of ​​the triangle from the rectangle and we have the area of ​​the figure: 800 - 16 = 784m^2

3 0
2 years ago
What is the domain of
ddd [48]
All real numbers is the correct answer
4 0
3 years ago
for some real number A, the graph of the line Y=(A+1)x+8 in the standard (x,y) coordinate plane passes through (2,6). What is th
kondaur [170]

Answer:

  -1

Step-by-step explanation:

The given equation has a y-intercept of 8, so goes through the point (0, 8).

If the line also goes through the point (2, 6) it has a slope of ...

  (y2 -y1)/(x2 -x1) = (6 -8)/(2 -0) = -2/2 = -1

The slope of the line is -1.

_____

The value of A is -2.

8 0
3 years ago
Sorting through unsolicited e-mail and spam affects the productivity of office workers. An InsightExpress survey monitored offic
nydimaria [60]

Answer:

1. Refer to the explanation for the frequency table.

2. (a) 60%

   (b) 25%

Step-by-step explanation:

Part 1. The question is asking us to group the data according to the classes given and fill in the frequency (F), relative frequency (RF), cumulative frequency (CF) and relative cumulative frequency (RCF).

To compute the frequency for each class count the number of data that lies in the class range and write it. For the first class (1-5) we can see that there are 12 numbers which lie in the range 1 to 5. Those numbers are: 2, 4, 4, 1, 2, 1, 5, 5, 5, 3, 4, 4. Similarly, for the second class, the frequency is 3 because 8, 8, 8 lie in this range from our existing data. For the third class 11-15 there are 2 numbers in the given data which lie in this range and those numbers are 12, 15. Similarly, the rest of the frequencies can be computed.

For the relative frequency, the frequency of each class is divided by the total frequency i.e. 20.  

For the class 1-5, RF = 12/20 = 0.6.  

For 6-10, RF = 3/20 = 0.15

For 11-15, RF = 2/20 = 0.1

For 16-20, RF = 1/20 = 0.05

For 21-25, RF=1/20 = 0.05

For 26-30, RF = 0/20 = 0.00

For 31-34, RF= 1/20 = 0.05.

To compute the cumulative frequency, add the existing frequency of each class with the previous frequencies.

For 1-5, CF = 0+12 = 12

For 6-10, CF = 12+3=15

For 11-15, CF = 12+3+2 = 17

For 16-20, CF=12+3+2+1 = 18

For 21-25, CF = 12+3+2+1+1=19

For 25-30, CF = 12+3+2+1+1+0=19

For 31-34, CF = 12+3+2+1+1+0+1=20

Now, to compute relative cumulative frequency, add the existing relative frequency of each class with the previous relative frequencies.  

For 1-5, RCF = 0+0.6

For 6-10, RCF = 0.6 + 0.15 = 0.75

For 11-15, RCF = 0.6+0.15+0.1 = 0.85

The rest of the relative cumulative frequencies can be computed in the same way.

Class(Minutes)           F       RF     CF     RCF

      1-5                       12     0.60    12     0.60  

      6-10                      3     0.15     15     0.75  

      11-15                      2     0.10     17     0.85  

      16-20                    1      0.05    18     0.90  

      21-25                    1      0.05    19     0.95  

      26-30                   0     0.00    19     0.95  

      31-34                     1     0.05    20       1  

       <u>Total                  20</u>  

Part 2. Now, we are asked to compute the Ogive graph which is also called as the cumulative frequency graph. The cumulative frequency needs to be plotted on the y-axis and the upper limit of each class needs to be plotted on the x-axis. The graph is attached.

(a) From the graph we can see that the number of workers who spend less than 5 minutes on unsolicited e-mail and spam are 12. So the answer for this part is 12 workers.

Percentage = 12/20 x 100 = 60%

(b) From the graph we can see that the number of workers who spend less than 10 minutes on spam e-mail are 15. The question is asking for the number of people who spend more than 10 minutes. For this we need to subtract 15 from the total number of workers.  

Number of workers spending more than 10 minutes = 20-15 = 5 workers.

Percentage = 5/20 x 100 = 25%

4 0
3 years ago
Determine if the following system of equations has no solutions, infinitely many
mylen [45]
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8 0
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