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Olenka [21]
3 years ago
12

Which of the following constants can be added to x2 - 3x to form a perfect square trinomial?

Mathematics
2 answers:
jeka57 [31]3 years ago
5 0

<em>Answer:</em>

<em>2 whole 1/4</em>

<em>Step-by-step explanation:</em>

<em>When forming a perfect square trinomial you need to "complete the square". </em>

<em>All of the steps to completing the square when solving an equation: </em>

<em>1.  The leading coefficient must be 1.  </em>

<em>2.  Divide b by 2. </em>

<em>3.  Square (b/2) </em>

<em>4.  Add (b/2)^2 to both sides to keep the polynomial balanced. </em>

<em>5.  You can now write the perfect square trinomial and solve. </em>

<em><u> </u></em>

x^2 - 3x  

-3/2

(-3/2)^2 = 9/4 = 2 1/4

hjlf3 years ago
5 0

Answer:

The answer is 2 1/4

Step-by-step explanation:

You are welcome.

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Need answer quickly! thank you in advance!
anyanavicka [17]

Answer:

b)(b²-a²)

Step-by-step explanation:

a cotθ + b cosecθ =p

b cotθ + a cosecθ =q

Now,

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={a(cotθ+cosecθ)+b(cotθ+cosecθ)} {a (cotθ-cosecθ)+b (cosecθ-cotθ)}

={a(cotθ+cosecθ)+b(cotθ+cosecθ)} [a (cotθ-cosecθ) + {- b (cotθ-cosecθ)} ]

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={(cotθ+cosecθ)(a+b)} {(cotθ-cosecθ) (a-b)}

=(cotθ+cosecθ) (a+b) (cotθ-cosecθ) (a-b)

=(cotθ+cosecθ) (cotθ-cosecθ) (a+b) (a-b)        

= (cot²θ-cosec²θ) (a²-b²)                                 [(a+b) (a-b)= (a²-b²)]

= -1 . (a²-b²)                               [ 1+cot²θ=cosec²θ ; ∴cot²θ-cosec²θ=-1]

=(b²-a²)

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