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den301095 [7]
3 years ago
5

Question is down below

Mathematics
2 answers:
mr Goodwill [35]3 years ago
7 0
The correct answer would be B.
True [87]3 years ago
6 0
As it is translated 5 units in right direction, all the x-coordinates will increase by 5, and as it's translated 4 units in upper direction, all the y-coordinates will increase by 4.

So, It's Coordinates will be: J'(15,9) K'(7,12) L'(9,13)

So, option B is your correct answer.

Hope it helped!
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Csc α - sin α =cos α cot α<br> Show how this is an identity?
Savatey [412]
Csc a - sin a = cos a cot a
1/sin a - sin a = cos a (cos a / sin a)
(1 - sin^2 a)/sin a = cos^2 a/sin a
cos^2 a/sin a = cos^2 a/sin a

Therefore, the given equation is an identity.
8 0
3 years ago
PLEASE SHOW YOUR WORK!
storchak [24]
Why isn't it have height width and lenght <span />
4 0
3 years ago
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
I know the answer, I'm just confused on what to put for explore, plan, solve, and examine!
mestny [16]

Answer:

Explore - ??? (What are you suppose to put for that (Add comment))

Plan - First week - 53 Second week - 62

Solve - 53 + 62 = 115

Examine - (Use a graph to see the rise?)

Step-by-step explanation:

5 0
3 years ago
Use the Change of Base Formula to evaluate log5 92. Then convert log5 92 to a logarithm in base 3. Round to the nearest thousand
ANTONII [103]
I was thinking that the answer  could be 

2.810; log3 21.903
         
            OR


<span> 2.810; log3 55.2 </span>

So,<span> log5 of 92 is between 2.5 and 3  Correct answer is A</span>

6 0
3 years ago
Read 2 more answers
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