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ser-zykov [4K]
3 years ago
8

Is the tangent ratio used for acute, obtuse, right or all?

Mathematics
1 answer:
sveta [45]3 years ago
8 0
The tangent ratio is used for right triangles
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The table show the age in years of employees in a company
adelina 88 [10]

Answer:

A. 24 ≤ a < 26.

B. 22.5

Step-by-step explanation:

A. Determination of the modal class interval.

Mode is the class with the highest frequency.

From the table given above, the highest frequency is 8, therefore the class will the highest frequency is:

24 ≤ a < 26.

B. To obtain the mean, we must determine the class mark. This is illustrated below:

Class >>>>> class mark >>> frequency

18 – 19 >>>> 18.5 >>>>>>>>> 3

20 – 21 >>> 20.5 >>>>>>>> 2

22 – 23 >>> 22.5 >>>>>>>> 7

24 – 25 >>> 24.5 >>>>>>>> 8

26 >>>>>>>> 26 >>>>>>>>> 0

The mean is given by the summation of the product of the class mark and frequency divided by the total frequency. This is illustrated below:

Mean = [(18.5x3) + (20.5x2) + (22.5x7) + (24.5x8) + (26x0)] / (3+2+7+8+0)

Mean = (55.5 + 41 + 157.5 + 196 + 0)/20

Mean = 450/20

Mean = 22.5

Therefore, the mean age is 22.5

4 0
2 years ago
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sleet_krkn [62]

Answer:

id say c. because its the closest number that corresponds with the graph.

5 0
3 years ago
What is the solution to the system of equations below? Negative 4 x 6 y = negative 18 and y = negative 2 x 21 (9, 3) (3, 9) (–9,
Firlakuza [10]

Answer:

start with root and minus with 3 and 8 and answer the following

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bogdanovich [222]
Problem 1)

AC is only perpendicular to EF if angle ADE is 90 degrees

(angle ADE) + (angle DAE) + (angle AED) = 180
(angle ADE) + (44) + (48) = 180
(angle ADE) + 92 = 180
(angle ADE) + 92 - 92 = 180 - 92
angle ADE  = 88

Since angle ADE is actually 88 degrees, we do NOT have a right angle so we do NOT have a right triangle

Triangle AED is acute (all 3 angles are less than 90 degrees)

So because angle ADE is NOT 90 degrees, this means AC is NOT perpendicular to EF

-------------------------------------------------------------

Problem 2)

a) The center is (2,-3) 

The center is (h,k) and we can see that h = 2 and k = -3. It might help to write (x-2)^2+(y+3)^2 = 9 into (x-2)^2+(y-(-3))^2 = 3^3 then compare it to (x-h)^2 + (y-k)^2 = r^2

---------------------

b) The radius is 3 and the diameter is 6

From part a), we have (x-2)^2+(y-(-3))^2 = 3^3 matching (x-h)^2 + (y-k)^2 = r^2

where
h = 2
k = -3
r = 3

so, radius = r = 3
diameter = d = 2*r = 2*3 = 6

---------------------

c) The graph is shown in the image attachment. It is a circle with center point C = (2,-3) and radius r = 3.

Some points on the circle are

A = (2, 0)
B = (5, -3)
D = (2, -6)
E = (-1, -3)

Note how the distance from the center C to some point on the circle, say point B, is 3 units. In other words segment BC = 3.

6 0
3 years ago
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