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klasskru [66]
3 years ago
11

Least common denominator for 1/49 1/21 1/3

Mathematics
2 answers:
Arte-miy333 [17]3 years ago
7 0
The LCD is 147 for this set of numbers.
I hope this helped!
Scorpion4ik [409]3 years ago
7 0
The least common denominator or lcd is 147 I hope this helps u pls thank and make brainliest
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Explain how to plot a point when given the coordinates.
velikii [3]
Here's the answer right here

7 0
3 years ago
Which of the following is being constructed in the image.
Sliva [168]

Answer:

  3. a line perpendicular to a given line through a point not on the line

Step-by-step explanation:

The point not on the line suggests that choices 1 and 3 are possibilities. The fact that the dotted line is not parallel (and is perpendicular) to the solid line suggests that choice 3 is applicable and choice 1 is not.

The short arcs are equidistant from the end points of the chord that intercepts the larger arc. Hence the line through the crossing point of the short arcs and the point on the other side of the line will be the perpendicular bisector of the chord, and will be perpendicular to the solid line. Creating that perpendicular is likely the purpose of the construction.

5 0
3 years ago
If tanA=a <br>then find sin4A-2sin2A/ sin4A+2sin2A​
anygoal [31]

Answer:

The value of the given expression is

\frac{sin4A-2sin2A}{sin4A+2sin2A}=-a^2

Step by step Explanation:

Given that tanA=a

To find the value of \frac{sin4A-2sin2A}{sin4A+2sin2A}

Let us find the value of the expression :

\frac{sin4A-2sin2A}{sin4A+2sin2A}=\frac{2cos2Asin2A-2sin2A}{2cos2Asin2A+2sin2A} ( by using the formula sin2A=2cosAsinA here A=2A)  

=\frac{2sin2A(cos2A-1)}{2sin2A(cos2A+1)}

=\frac{(cos2A-1)}{(cos2A+1)}

=\frac{(-(1-cos2A))}{(1+cos2A)}(using  sin^2A+cos^2A=1  here A=2A)

=\frac{-(sin^2A+cos^2A-(cos^2A-sin^2A))}{sin^2A+cos^2A+(cos^2A-sin^2A)}(using cos2A=cos^2A-sin^2A here A=2A)

=\frac{-(sin^2A+cos^2A-cos^2A+sin^2A)}{sin^2A+cos^2A+(cos^2A-sin^2A)}

=\frac{-(sin^2A+sin^2A)}{cos^2A+cos^2A}

=\frac{-2sin^2A}{2cos^2A}

=-\frac{sin^2A}{cos^2A}

=-tan^2A  ( using tanA=\frac{sinA}{cosA} here A=2A )

 =-a^2 (since tanA=a given )

Therefore \frac{sin4A-2sin2A}{sin4A+2sin2A}=-a^2

6 0
3 years ago
Pls??? I’ll help u if u help me!!!
il63 [147K]

Answer: the answer is 21

Step-by-step explanation:

5 0
3 years ago
What’s the missing side length 28 21
Sergio039 [100]

Answer:

HOW ON THE WORLD WE SUPPOSED TO ANSWER DIS ?!?!?!

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
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