Solve by using tables. give each answer to at most two decimal places. –2x2 – 4 = –8x (1 point) 0.59, 3.41 –3.41, –0.59 1.17, 6.83 –1.41, 1.41

x=3.414 and x=0.59 is the desired answer.

Step-by-step explanation:

We are given a quadratic equation as :

on dividing both side by 2 we have the equation as

on solving for the roots of the equation using the quadratic formula we get that

and

x=2+1.414 and x=2-1.414

Hence , x=3.414 and x=0.586≈0.59.

Answer

0.59, 3.41

Explanation

–2x2 – 4 = –8x is a quadrilatic equation

–2x2 – 4 = –8x  ⇒ -2x² + 8x - 4 = 0

                                2x² - 8x + 4 = 0

                                 x² - 4x + 2 = 0

Using the complete square method to solve;

x² - 4x = -2

x² - 4x + 2² = -2 + 2²

(x - 2)² = 2

x - 2 = √2

x = √2 + 2

∴ x = 1.414 + 2             or  x = -1.414 + 2

  x = 3.414                          x = 0.5858

     = 3.41                                  = 0.59

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.

answer: your gay

step-by-step explanation:

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