Derive the equation of the parabola with a focus at (-5, -5) and a directrix of y = 7

Step-by-step explanation:

When we plot the focus and draw in the directrix, there are 12 units between them.  Directly between them will sit the vertex.  Since the parabola by definition will wrap itself around the focus, we know that this is an upside down parabola with the vertex aligning itself with the x coordinate of the focus.  So the vertex is at (-5, 1) since that is the point that is in line with x coordinate of the focus and the y coordinate of 1 is halfway between the focus and the directrix.  The form for this type of parabola that opens upside down is:

,

where p is the distance between the vertex and the focus (or the vertex and the directrix since it is the same distance), h is the first coordinate of the vertex and k is the second coordinate of the vertex.

We need to get those numbers filled in there and then put the parabola either into work form or in standard form.  

We know from up above that there are 12 units between the focus and the directrix, we know that the vertex is at (-5, 1), so that means that p is the number of units between the vertex and the focus which is 6.  We have everything we need now to fill in our equation.

If we put this into vertex form, we need to simplify the above equation into

so we will do some simplifying on our equation.

and

.  One last step is to add 1 to both sides:

∠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.

24

step-by-step explanation:

the graph shows 30% of students walk. 80 students were questioned, so the number that said they walk is

30% × 80 = 0.30 × 80 = 24