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Greeley [361]
3 years ago
10

Solve each equation. I don't know this

Mathematics
1 answer:
MArishka [77]3 years ago
5 0
1.(2x - 3)(x + 7) = 0
   2x^{2} + 14x - 3x - 21 = 0                                  
   2x^{2} + 11x - 21 = 0                       
  
   x = \frac{-11 +/- \sqrt{11^{2} - 4(2)(-21)}}{2(2)}                 
  
   x = \frac{-11 +/- \sqrt{121 + 168}}{4}
  
   x = \frac{-11 +/- \sqrt{289}}{4}
  
   x = \frac{-11 +/- 17}{4}
  
   x = -2.75 +/- 4.25
   x = -2.75 + 4.25                     x = -2.75 - 4.25
   x = 1.5                                                   <u></u>x = -7
----------------------------------------------------------------------------------------------------------  2.8x(2x - 5) = 0
   8x(2x) - 8x(5) = 0
   16x^{2} - 40x = 0
   16x^{2} - 4x + 0 = 0
  
   x = \frac{-(-40) +/- \sqrt{(-40)^{2} - 4(16)(0)}}{2(16)}
  
   x = \frac{40 +/- \sqrt{1600 - 0}}{32}
  
   x = \frac{40 +/- \sqrt{1600}}{32}
  
   x = \frac{40 +/- 40}{32}
  
   x = 1.25 +/- 1.25
   x = 1.25 + 1.25                                  x = 1.25 - 1.25
   x = 2.5                                               x = 0
----------------------------------------------------------------------------------------------------------
3.x^{2} + 3x - 10 = 0
  
   x = \frac{-3 +/- \sqrt{3^{2} - 4(1)(-10)}}{2(1)}
  
   x = \frac{-3 +/- \sqrt{9 + 40}}{2}
  
   x = \frac{-3 +/- \sqrt{49}}{2}
  
   x = \frac{-3 +/- {7}}{2}
  
   x = -1.5 +/- 3.5
   x = -1.5 + 3.5               x = -1.5 - 3.5
   x = 2                             x = -5
----------------------------------------------------------------------------------------------------------
4. x^{2} = 13x - 36
    x^{2} - 13x + 36 = 13x - 13x - 36 + 36
    x^{2} - 13x + 36 = 0
   
    x = \frac{-(-13) +/- \sqrt{(-13)^{2} - 4(1)(36)}}{2(1)}

   
    x = \frac{13 +/- \sqrt{169 - 144}}{2}
  
    x = \frac{13 +/- \sqrt{25}}{2}
 
    x = \frac{13 +/- 5}{2}

    x = 6.5 +/- 2.5
    x = 6.5 + 2.5                     x = 6.5 - 2.5
    x = 9                                 x = 4
----------------------------------------------------------------------------------------------------------
5.3x^{2} - 7x + 2 = 0
  
   x = \frac{-(-7) +/- \sqrt{(-7)^{2} - 4(1)(2)}}{2(3)}

   x = \frac{7 +/- \sqrt{49 - 8}}{6}

   x = \frac{7 +/- \sqrt{41}}{6}

   x = \frac{7 +/- 6.403124237432849}{6}
 
   x = 1.167+/- 1.06718737290547
   x = 1.67 + 1.06718737290547              x = 1.167 - 1.06718737290547
   x = 2.73718737290547                         x = 0.60281262709453
----------------------------------------------------------------------------------------------------------
6.10x^{2} - 10x + 9 = 5x^{2} + 4x + 1
   10x^{2} - 5x^{2} - 10x + 10x + 9 - 1 = 5x^{2} - 5x^{2} + 4x + 10x + 1 - 1
   5x^{2} + 8 = 14x
   5x^{2} - 14x + 8 = 14x - 14x
   <u />5x^{2} - 14x + 8 = 0
  
   x = \frac{-(-14) + \sqrt{(-14)^{2} - 4(5)(8)}}{2(5)}
  
   x = \frac{14 +/- \sqrt{196 - 160}}{10}
  
   x = \frac{14 +/- \sqrt{36}}{10}.
  
   x = \frac{14 +/- 6}{10}
  
   x = 1.4 +/- 0.6
   x = 1.4 + 0.6                x = 1.4 - 0.6
   <u />x = 2                            x = 0.8
 
 
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Find the discriminant, describe the types of roots, and find the solution for 3x^2-24x+12
solniwko [45]
The discriminant of a polynomial is given by:
 b ^ 2-4ac
 Substituting the values we have:
 (-24) ^ 2-4 * (3) * (12) = 432
 Since the discriminator is greater than zero, then the roots are real.
 x = (- b +/- root (b ^ 2-4ac)) / (2a)
 Substituting the values:
 x = (- (- 24) +/- root (432) / (2 * (3))
 x = (- (- 24) +/- root (432) / (2 * (3))
 x = (- (- 24) +/- root (144 * 3) / (2 * (3))
 x = (24 +/- 12raiz (3) / (6)
 x = 4 +/- 2raiz (3)
 The roots are:
 x1 = 4 + 2raiz (3)
 x2 = 4 - 2raiz (3)
 Answer: 
 432
 the roots are real.
 
x1 = 4 + 2raiz (3)
 
x2 = 4 - 2raiz (3)
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There is a parallelogram ABCD with diagonals AC and BD. The diagonals AC and BD intersects each other at point E. Side AB is con
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Answer:

SAS theorem

Step-by-step explanation:

Given

\square ABCD

\[ \lvert \[ \lvert AB =\[ \lvert \[ \lvert CD

\angle BAC = \angle  DCA

Required

Which theorem shows △ABE ≅ △CDE.

From the question, we understand that:

AC and BD intersects at E.

This implies that:

\[ \lvert \[ \lvert AE = \[ \lvert \[ \lvert EC

and

\[ \lvert \[ \lvert BE = \[ \lvert \[ \lvert ED

So, the congruent sides and angles of △ABE and △CDE are:

\[ \lvert \[ \lvert AB =\[ \lvert \[ \lvert CD ---- S

\angle BAC = \angle  DCA ---- A

\[ \lvert \[ \lvert BE = \[ \lvert \[ \lvert ED or \[ \lvert \[ \lvert AE = \[ \lvert \[ \lvert EC  --- S

<em>Hence, the theorem that compares both triangles is the SAS theorem</em>

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