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dexar [7]
2 years ago
12

Can you factor 25m^2-10mn+n^2

Mathematics
1 answer:
Charra [1.4K]2 years ago
4 0
You can factor it into (-5m+n)(-5m+n)
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Find all solutions of the equation: 2cos^2x-cosx=1
Art [367]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/3166243

——————————

Solve the trigonometric equation:

     \mathsf{2\,cos^2\,x-cos\,x=1}\\\\ \mathsf{2\,cos^2\,x-cos\,x-1=0}


Make a substitution:

     \mathsf{cos\,x=t\qquad (-1\le t\le 1)}

and the equation becomes

     \mathsf{2t^2-t-1=0}


Rewrite conveniently  – t  as  + t – 2t,  and then factor the left-hand side by grouping:

      \mathsf{2t^2+t-2t-1=0}\\\\ \mathsf{t\cdot (2t+1)-1\cdot (2t+1)=0}


Factor out  2t + 1:

     \mathsf{(2t+1)\cdot (t-1)=0}\\\\ \begin{array}{rcl} \mathsf{2t+1=0}&~\textsf{ or }~&\mathsf{t-1=0}\\\\ \mathsf{2t=1}&~\textsf{ or }~&\mathsf{t=1}\\\\ \mathsf{t=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{t=1} \end{array}


Substitute back for  t = cos x:

     \begin{array}{rcl}\mathsf{cos\,x=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{cos\,x=1}\\\\ \mathsf{cos\,x=cos\,60^\circ}&~\textsf{ or }~&\mathsf{cos\,x=cos\,0} \end{array}


Therefore,

     \begin{array}{rcl} \mathsf{x=\pm\,60^\circ+k\cdot 360^\circ}&~\textsf{ or }~&\mathsf{cos\,x=0+k\cdot 360^\circ} \end{array}

where  k  is an integer.


Solution set:   

\mathsf{S=\left\{x\in\mathbb{R}:~~x=-\,60^\circ+k\cdot 360^\circ~~or~~x=60^\circ+k\cdot 360^\circ~~or~~x=k\cdot 360^\circ,~~k\in\mathbb{Z}\right\}}


I hope this helps. =)

3 0
3 years ago
Translate triangle ABC 2 units left and 3 units up
kifflom [539]
1) Point A will be at (-4,4)
2) I think an equilateral triangle can't be right triangle
7 0
3 years ago
Read 2 more answers
What is the median of 40,33,37,54,41,34,27,39,35
Sauron [17]

Answer:

It's 37

Step-by-step explanation:

The given data set is ;

40,33,37,54,41,34,27,39,35

We arrange the data set in ascending order to obtain;

{27,33,34,35,37,39,40,41,54}

The median is the middle number {37}

8 0
3 years ago
If JM = 5x – 8 and LM = 2x - 6, which expression represents JL?
Anastasy [175]

Answer:

3x-2

Step-by-step explanation:

JK =JL+LM

5x-8=JL +(2x -6)

JL = 5x-8 -2x+6

JL = 3x-2

6 0
3 years ago
What are the like terms of 10a+4b+3a-15b
TiliK225 [7]
The like terms in 10a+4b+3a-15b consist of a and b. First, we have to combine like terms.
(10a+3a) + (4b-15b)
solve a and b
13a - 11b. Thus, the like terms in your original expression are 13a and -11b :)
3 0
3 years ago
Read 2 more answers
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