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Serhud [2]
3 years ago
8

Which graph shows a function whose inverse is also a function?

Mathematics
1 answer:
Westkost [7]3 years ago
7 0

Answer:

hello the function graph is turned 90 degree like clock way.

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M(6,5) is the mid-point of the straight line joining A(2 , 3) to the
charle [14.2K]

Given :

M(6,5) is the mid-point of the straight line joining A(2 , 3) to the

point B.

To Find :

The coordinates of B.

Solution :

Let, coordinates of point B is ( h , k ) .

It is given that M is the mid - point of the straight line joining points A and B.

Coordinates of M is given by :

M = (\dfrac{2+h}{2},\dfrac{3+k}{2})\\\\(6,5 ) = (\dfrac{2+h}{2},\dfrac{3+k}{2})\\\\\dfrac{2+h}{2} = 6 \ , \  \dfrac{3+k}{2} = 5\\\\h = 10 \ ,\ k = 7

Therefore, coordinates of point B is ( 10 , 7 ).

7 0
2 years ago
PLEASE HELP!!
slava [35]

Answer:

99   (cm^2)

Step-by-step explanation:

Perpendicular to the AB segment at points D and C, the graph is divided into two triangles and a rectangle.

The area of the middle rectangle is equal to 8*9=72. The hypotenuse of the right triangle is 10cm, and one of the right sides is 9cm, so the other side is SQRT (10^2-9^2) = SQRT (19).

One side of the left triangle is 9cm long and the other side is 14-8-sqRT (19) = 6-sqRT (19) cm.

Then, add the area of the three parts.

72+9*sqrt(19)/2+9*(6-sqrt(19))/2=99    (cm^2)

4 0
3 years ago
Which Statement about 4(x-3) is True?
Sophie [7]

Answer: it’s b because if you times that by anything it will always be a product I think

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
6,300,000,000 = 6.3 x 1,000,000,000
Nonamiya [84]
No they are not

scientific notation requires writing the ' × 10^9' part as a 10 with an exponent
7 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
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