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Marrrta [24]
3 years ago
5

Given right triangle DEF, what is the value of tan(F) 9/41 40/41 40/9  41/9

Mathematics
2 answers:
Rom4ik [11]3 years ago
7 0
Check the picture below.

11Alexandr11 [23.1K]3 years ago
4 0
The answer is C.) 40/9 

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Does a point have a length and a width
zaharov [31]

Answer:

The answer is no.

Step-by-step explanation:

A point is neither a line or a ray, or anything that involves a line. It's just a dot, a location.

4 0
3 years ago
Read 2 more answers
Use Euclid's division algorithm to find the HCF of 441, 567, 693
Illusion [34]
Let a = 693, b = 567 and c = 441

Now first we will find HCF of 693 and 567 by using Euclid’s division algorithm as under

693 = 567 x 1 + 126
567 = 126 x 4 + 63
126 = 63 x 2 + 0
Hence, HCF of 693 and 567 is 63

Now we will find HCF of third number i.e., 441 with 63 So by Euclid’s division alogorithm for 441 and 63

441 = 63 x 7+0
=> HCF of 441 and 63 is 63.

Hence, HCF of 441, 567 and 693 is 63.
6 0
3 years ago
Triangle A E C is shown. Line segment B D is drawn near point C to form triangle B D C.
AVprozaik [17]

∠BDC and ∠AED are right angles, is  a piece of additional information is appropriate to prove △ CEA ~ △ CDB

Triangle AEC is shown. Line segment B, D is drawn near point C to form triangle BDC.

<h3> What are Similar triangles?</h3>

Similar triangles, are those triangles which have similar properties,i.e. angles and proportionality of sides.

Image is attached below,
as shown in figure
∡ACE = ∡BCD ( common angle )
∡AED = ∡BDC ( since AE and BD are perpendicular to same line EC and make right angles as E and C)
∡EAC =- ∡DBC ( corresponding angles because AE and BD are parallel lines)

Thus, △CEA ~ △CDB , because of the two perpendiculars AE and BD.

Learn more about similar triangles here:
brainly.com/question/25882965

#SPJ1


3 0
1 year ago
In 2013, voyager 1 traveled 1,468,800 kilometers each day. Whats that in scientific notation
n200080 [17]

Answer: 1.4688 x 10^6

Step-by-step explanation: Number has to be in between 1 and 10. Moved the decimal 6 places so that is your exponent

8 0
2 years ago
Please help!! <br>circles combos are too hard​
malfutka [58]

Answer:

56 + 53pi

Step-by-step explanation:

<u><em>Area of small circles:</em></u>

diameter of small circle: 4cm

forumla to find area of circle: A = pir^2

r is radius = half of diameter -> d/2 = 4 / 2 = 2cm

A = pi (2cm)^2

A = pi (4cm)

A = 4pi

<u><em>Area of large circle:</em></u>

diameter of small circle: 4cm

forumla to find area of circle: A = pir^2

r is radius = half of diameter -> d/2 = 14 / 2 = 7cm

A = pi (7cm)^2

A = pi (49cm)

A = 49pi

<u><em>Area of rectangle:</em></u>

Area = width x length

Area = 14cm x 4cm

Area = 56cm

<u><em>Add all three areas:</em></u>

Area of rectangle + large circle + small circle

56cm + 49pi + 4pi = 56cm + 53pi

7 0
2 years ago
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