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elixir [45]
2 years ago
5

Help me with which triangle it is and how do I find x?

Mathematics
2 answers:
weeeeeb [17]2 years ago
7 0

Step-by-step explanation:

<em>it's </em><em>an </em><em>isosceles</em><em> triangle</em><em>,</em>

<em>as </em><em>an </em><em>Isosceles</em><em> triangle</em><em> </em><em>have </em><em>two </em><em>equal</em><em> </em><em>sides </em><em>so</em>

<em>→</em><em> </em><em>2</em><em>x</em><em> </em><em>-</em><em> </em><em>9</em><em> </em><em>=</em><em> </em><em>x </em><em>+</em><em> </em><em>5</em>

<em>→</em><em> </em><em>2</em><em>x</em><em> </em><em>-</em><em> </em><em>x</em><em> </em><em>=</em><em> </em><em>5</em><em> </em><em>+</em><em> </em><em>9</em>

<em>→</em><em> </em><em> </em><em>x </em><em>=</em><em> </em><em>1</em><em>4</em>

<em>therefore</em><em>,</em><em> </em><em>value</em><em> </em><em>of </em><em>x </em><em>is </em><em>1</em><em>4</em><em>.</em>

<em>2</em><em>(</em><em>1</em><em>4</em><em>)</em><em> </em><em>-</em><em> </em><em>9</em><em> </em><em>=</em><em> </em><em>2</em><em>8</em><em> </em><em>-</em><em> </em><em>9</em><em> </em><em>=</em><em> </em><em>1</em><em>9</em>

<em>1</em><em>4</em><em> </em><em>+</em><em> </em><em>5</em><em> </em><em>=</em><em> </em><em>1</em><em>9</em>

<em>hope </em><em>this</em><em> answer</em><em> helps</em><em> you</em><em> dear</em><em> take</em><em> care</em><em>!</em>

3241004551 [841]2 years ago
6 0

Answer:

isosceles

x=14

Step-by-step explanation:

isosceles because of the two I symbols on the triangle

x+5=2x-9

5+9=2x-x

14=x

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The ratio of the measures of the acute angles of a right triangle is $8:1$. In degrees, what is the measure of the largest angle
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Answer:

The angles in the triangle and 10, 80 and 90

Step-by-step explanation:

The two angles that are left in a right triangle add to 90 degrees

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5x minus 4 equals x squared minus 4x plus 4. What is x
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Two solutions were found :

x =(4-√-64)/-10=2/-5+4i/5= -0.4000-0.8000i

x =(4+√-64)/-10=2/-5-4i/5= -0.4000+0.8000i

Step by step solution :

Step  1  :

Equation at the end of step  1  :

 ((0 -  5x2) -  4x) -  4  = 0

Step  2  :

Step  3  :

Pulling out like terms :

3.1     Pull out like factors :

  -5x2 - 4x - 4  =   -1 • (5x2 + 4x + 4)

Trying to factor by splitting the middle term

3.2     Factoring  5x2 + 4x + 4

The first term is,  5x2  its coefficient is  5 .

The middle term is,  +4x  its coefficient is  4 .

The last term, "the constant", is  +4

Step-1 : Multiply the coefficient of the first term by the constant   5 • 4 = 20

Step-2 : Find two factors of  20  whose sum equals the coefficient of the middle term, which is   4 .

Observation : No two such factors can be found !!

Conclusion : Trinomial can not be factored

Equation at the end of step  3  :

 -5x2 - 4x - 4  = 0

Step  4  :

Parabola, Finding the Vertex :

4.1      Find the Vertex of   y = -5x2-4x-4

For any parabola,Ax2+Bx+C,the  x -coordinate of the vertex is given by  -B/(2A) . In our case the  x  coordinate is  -0.4000  

Plugging into the parabola formula  -0.4000  for  x  we can calculate the  y -coordinate :

 y = -5.0 * -0.40 * -0.40 - 4.0 * -0.40 - 4.0

or   y = -3.200

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = -5x2-4x-4

Axis of Symmetry (dashed)  {x}={-0.40}

Vertex at  {x,y} = {-0.40,-3.20}

Function has no real roots

Solve Quadratic Equation by Completing The Square

4.2     Solving   -5x2-4x-4 = 0 by Completing The Square .

Multiply both sides of the equation by  (-1)  to obtain positive coefficient for the first term:

5x2+4x+4 = 0  Divide both sides of the equation by  5  to have 1 as the coefficient of the first term :

  x2+(4/5)x+(4/5) = 0

Subtract  4/5  from both side of the equation :

  x2+(4/5)x = -4/5

Add  4/25  to both sides of the equation :

 On the right hand side we have :

  -4/5  +  4/25   The common denominator of the two fractions is  25   Adding  (-20/25)+(4/25)  gives  -16/25

 So adding to both sides we finally get :

  x2+(4/5)x+(4/25) = -16/25

Adding  4/25  has completed the left hand side into a perfect square :

  x2+(4/5)x+(4/25)  =

  (x+(2/5)) • (x+(2/5))  =

 (x+(2/5))2

Things which are equal to the same thing are also equal to one another. Since

  x2+(4/5)x+(4/25) = -16/25 and

  x2+(4/5)x+(4/25) = (x+(2/5))2

then, according to the law of transitivity,

  (x+(2/5))2 = -16/25

Note that the square root of

  (x+(2/5))2   is

  (x+(2/5))2/2 =

 (x+(2/5))1 =

  x+(2/5)

Now, applying the Square Root Principle to  Eq. #4.2.1  we get:

  x+(2/5) = √ -16/25

Subtract  2/5  from both sides to obtain:

  x = -2/5 + √ -16/25

Since a square root has two values, one positive and the other negative

  x2 + (4/5)x + (4/5) = 0

  has two solutions:

 x = -2/5 + √ 16/25 •  i

  or

 x = -2/5 - √ 16/25 •  i

Note that  √ 16/25 can be written as

 √ 16  / √ 25   which is 4 / 5

Solve Quadratic Equation using the Quadratic Formula

4.3     Solving    -5x2-4x-4 = 0 by the Quadratic Formula .

According to the Quadratic Formula,  x  , the solution for   Ax2+Bx+C  = 0  , where  A, B  and  C  are numbers, often called coefficients, is given by :

                                   

           - B  ±  √ B2-4AC

 x =   ————————

                     2A

 In our case,  A   =     -5

                     B   =    -4

                     C   =   -4

Accordingly,  B2  -  4AC   =

                    16 - 80 =

                    -64

Applying the quadratic formula :

              4 ± √ -64

  x  =    —————

                   -10

In the set of real numbers, negative numbers do not have square roots. A new set of numbers, called complex, was invented so that negative numbers would have a square root. These numbers are written  (a+b*i)

Both   i   and   -i   are the square roots of minus 1

Accordingly,√ -64  =

                   √ 64 • (-1)  =

                   √ 64  • √ -1   =

                   ±  √ 64  • i

Can  √ 64 be simplified ?

Yes!   The prime factorization of  64   is

  2•2•2•2•2•2

To be able to remove something from under the radical, there have to be  2  instances of it (because we are taking a square i.e. second root).

√ 64   =  √ 2•2•2•2•2•2   =2•2•2•√ 1   =

               ±  8 • √ 1   =

               ±  8

So now we are looking at:

          x  =  ( 4 ± 8i ) / -10

Two imaginary solutions :

x =(4+√-64)/-10=2/-5-4i/5= -0.4000+0.8000i

 or:

x =(4-√-64)/-10=2/-5+4i/5= -0.4000-0.8000i

Two solutions were found :

x =(4-√-64)/-10=2/-5+4i/5= -0.4000-0.8000i

x =(4+√-64)/-10=2/-5-4i/5= -0.4000+0.8000i

<em>hope i helped</em>

<em>-Rin:)</em>

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