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MatroZZZ [7]
3 years ago
13

How do you factor this trinomial 4a^2-4a+1

Mathematics
1 answer:
ElenaW [278]3 years ago
4 0

Answer:

(2a-1)(2a-1)

Step-by-step explanation:

To write the factored form, multiply a*c from the standard form ax^2+bx+c. Here it is 4*1 = 4.

Then find factors of 4 that add to b=-4.

4: 1,2,4

-2+-2 = -4

Split the middle term into -2x+-2x and factor by grouping.

4a^2-2a+-2a+1\\(4a^2-2a)+(-2a+1)\\2a(2a-1)-1(2a-1)\\(2a-1)(2a-1)


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ludmilkaskok [199]
The answer is:  "3" .
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Use the Pythagorean theorem (for right triangles):

a² + b² = c² ;

in which "a = "side length 1" (unknown; for which we which to solve);

               "b" = "side length 2" = "√3" (given in the figure) ;

               "c" = "length of hypotenuse" = "2√3" (given in the figure);
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a² + b² = c² ; 

a² = c² − b² ;

Plug in the known values for "c" and "b" ;

a² = (2√3)²  −  (√3)² ;

Simplify:

(2√3)² = 2² * (√3)² = 2 * 2 * (√3√3)  = 4 * 3 =  12 .

 (√3)²  = (√3√3) = 3 .

a²  = 12 − 3 = 9 .

a² = 9

Take the "positive square root" of EACH SIDE of the equation; to isolate "a" on one side of the equation;  & to solve for "a" ; 

+√(a²)  = +√9 ;

     a = 3 . 
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The answer is:  "3" .
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3 0
3 years ago
Is UVW=Xyz?if so, name the postulate that applies
aivan3 [116]

By critically observing the two triangles, we can deduce that they: B. might not be congruent.

<h3>The properties of similar triangles.</h3>

In Geometry, two triangles are said to be similar when the ratio of their corresponding sides are equal in magnitude and their corresponding angles are congruent.

By critically observing the two triangles, we can logically deduce that the three angles of both triangles are congruent in accordance with AAA similarity postulate:

  • <U ≅ <X
  • <V ≅ <Y
  • <W ≅ <Z

However, AAA isn't a congruence postulate and as such all similar triangles might not be congruent.

Read more on congruency here: brainly.com/question/11844452

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2 years ago
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3 years ago
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Alenkasestr [34]
<h3>Given</h3>

tan(x)²·sin(x) = tan(x)²

<h3>Find</h3>

x on the interval [0, 2π)

<h3>Solution</h3>

Subtract the right side and factor. Then make use of the zero-product rule.

... tan(x)²·sin(x) -tan(x)² = 0

... tan(x)²·(sin(x) -1) = 0

This is an indeterminate form at x = π/2 and undefined at x = 3π/2. We can resolve the indeterminate form by using an identity for tan(x)²:

... tan(x)² = sin(x)²/cos(x)² = sin(x)²/(1 -sin(x)²)

Then our equation becomes

... sin(x)²·(sin(x) -1)/((1 -sin(x))(1 +sin(x))) = 0

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Another way to do it is to multiply top and bottom by 4
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Either way, the answer is 64%
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3 years ago
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