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Novosadov [1.4K]
2 years ago
11

How do you think we could use estimating irrational numbers?

Mathematics
1 answer:
bekas [8.4K]2 years ago
3 0

Answer:

An irrational number is a number that cannot be written as a fraction. It is a non-repeating, non-terminating decimal. Approximate square root of numbers that are not perfect squares and put them on the number line.

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Alex, Toby and Samuel are playing a game together.
Shalnov [3]

Answer:

1/9

Step-by-step explanation:

There are three places and three people.

Times them together for the total outcomes:

1/3 x 1/3= 1/9

8 0
3 years ago
MA = (4x - 2) and mzB = (11x + 17)°. Find x if the angles are
charle [14.2K]
The answer is a.) 165 sorry if that’s wrong but i hope it’s right
3 0
3 years ago
Read 2 more answers
Verify sine law by taking triangle in 4 quadrant<br>Explain with figure.<br>​
Ksivusya [100]

Proof of the Law of Sines

The Law of Sines states that for any triangle ABC, with sides a,b,c (see below)

a

 sin  A

=

b

 sin  B

=

c

 sin  C

For more see Law of Sines.

Acute triangles

Draw the altitude h from the vertex A of the triangle

From the definition of the sine function

 sin  B =

h

c

    a n d        sin  C =

h

b

or

h = c  sin  B     a n d       h = b  sin  C

Since they are both equal to h

c  sin  B = b  sin  C

Dividing through by sinB and then sinC

c

 sin  C

=

b

 sin  B

Repeat the above, this time with the altitude drawn from point B

Using a similar method it can be shown that in this case

c

 sin  C

=

a

 sin  A

Combining (4) and (5) :

a

 sin  A

=

b

 sin  B

=

c

 sin  C

- Q.E.D

Obtuse Triangles

The proof above requires that we draw two altitudes of the triangle. In the case of obtuse triangles, two of the altitudes are outside the triangle, so we need a slightly different proof. It uses one interior altitude as above, but also one exterior altitude.

First the interior altitude. This is the same as the proof for acute triangles above.

Draw the altitude h from the vertex A of the triangle

 sin  B =

h

c

      a n d          sin  C =

h

b

or

h = c  sin  B       a n d         h = b  sin  C

Since they are both equal to h

c  sin  B = b  sin  C

Dividing through by sinB and then sinC

c

 sin  C

=

b

 sin  B

Draw the second altitude h from B. This requires extending the side b:

The angles BAC and BAK are supplementary, so the sine of both are the same.

(see Supplementary angles trig identities)

Angle A is BAC, so

 sin  A =

h

c

or

h = c  sin  A

In the larger triangle CBK

 sin  C =

h

a

or

h = a  sin  C

From (6) and (7) since they are both equal to h

c  sin  A = a  sin  C

Dividing through by sinA then sinC:

a

 sin  A

=

c

 sin  C

Combining (4) and (9):

a

 sin  A

=

b

 sin  B

=

c

 sin  C

7 0
2 years ago
Is there an ordered pair solutions to <br> -3x+3y=4 and <br> -x+y=3 ????
elixir [45]
Y=x+3
-3x+3(x+3)=4
-3x+3x+9=4
9=4
Since this statement is false, there are no ordered pair solutions to this problem.
8 0
2 years ago
Which of the following is the conjugate of 10 – 3i?
ipn [44]

we know that

The conjugate of a complex number is the number with equal real part and imaginary part equal in magnitude but opposite in sign

so

In this problem we have

10-3i

the conjugate is equal to--------> 10+3i

therefore

<u>the answer is the option</u>

10+3i

3 0
2 years ago
Read 2 more answers
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