4x2 = 3x + 20
4x2= 3x+20

3.2     Factoring  4x2 + 3x - 20 

The first term is,  4x2  its coefficient is  4 .

The middle term is,  +3x  its coefficient is  3 .

The last term, "the constant", is  -20 

Step-1 : Multiply the coefficient of the first term by the constant   4 • -20 = -80 

Step-2 : Find two factors of  -80  whose sum equals the coefficient of the middle term, which is   3 .

     -80   +   1   =   -79     -40   +   2   =   -38     -20   +   4   =   -16     -16   +   5   =   -11     -10   +   8   =   -2     -8   +   10   =   2     -5   +   16   =   11     -4   +   20   =   16     -2   +   40   =   38     -1   +   80   =   79

Step-by-step explanation:

4.1      Find the Vertex of   y = -4x2-3x+20

4.1      Find the Vertex of   y = -4x2-3x+20Parabolas have a highest or a lowest point called the Vertex .   Our parabola opens down and accordingly has a highest point (AKA absolute maximum) .    We know this even before plotting  "y"  because the coefficient of the first term, -4 , is negative (smaller than zero). 

4.1      Find the Vertex of   y = -4x2-3x+20Parabolas have a highest or a lowest point called the Vertex .   Our parabola opens down and accordingly has a highest point (AKA absolute maximum) .    We know this even before plotting  "y"  because the coefficient of the first term, -4 , is negative (smaller than zero).  Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two  x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions. 

4.1      Find the Vertex of   y = -4x2-3x+20Parabolas have a highest or a lowest point called the Vertex .   Our parabola opens down and accordingly has a highest point (AKA absolute maximum) .    We know this even before plotting  "y"  because the coefficient of the first term, -4 , is negative (smaller than zero).  Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two  x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions.  Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex. 

step-by-step explanation:

∠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.