NEED ANSWER FAST WILL GIVE BRAINLIEST what is −3.9 − 8.9
1 answer:
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Answer:
Solution ( Second Attachment ) : - 2.017 + 0.656i
Solution ( First Attachment ) : 16.140 - 5.244i
Step-by-step explanation:
Second Attachment : The quotient of the two expressions would be the following,
÷
So if we want to determine this expression in standard complex form, we can first convert it into trigonometric form, then apply trivial identities. Either that, or we can straight away apply the following identities and substitute,
( 1 ) cos(x) = sin(π / 2 - x)
( 2 ) sin(x) = cos(π / 2 - x)
If cos(x) = sin(π / 2 - x), then cos(2π / 5) = sin(π / 2 - 2π / 5) = sin(π / 10). Respectively sin(2π / 5) = cos(π / 2 - 2π / 5) = cos(π / 10). Let's simplify sin(π / 10) and cos(π / 10) with two more identities,
( 1 )
( 2 )
These two identities makes sin(π / 10) = , and cos(π / 10) =
.
Therefore cos(2π / 5) = , and sin(2π / 5) =
. Substitute,
÷
Remember that cos(- π / 2) = 0, and sin(- π / 2) = - 1. Substituting those values,
÷
And now simplify this expression to receive our answer,
÷
=
,
=
and
=
=
As you can see our solution is option c. - 2.01749 was rounded to - 2.017, and 0.65552 was rounded to 0.656.
________________________________________
First Attachment : We know from the previous problem that cos(2π / 5) = , sin(2π / 5) =
, cos(- π / 2) = 0, and sin(- π / 2) = - 1. Substituting we receive a simplified expression,
We know that and
. Therefore,
Solution :
Which rounds to about option b.
So, notice, the focus point is at -7, 5, and the directrix is at y = -11.
keep in mind that the vertex is half-way between those two fellows, and the distance from the vertex to either one of them is "p" units, check the picture below.
with that focus point and that directrix, the half-way over the axis of symmetry will be -7, -3, that's where the vertex is at, and notice the distance "p", is 8 units.
since the parabola is opening upwards, "p" is positive 8.
![\bf \textit{parabola vertex form with focus point distance}\\\\ \begin{array}{llll} 4p(x- h)=(y- k)^2 \\\\ \boxed{4p(y- k)=(x- h)^2} \end{array} \qquad \begin{array}{llll} vertex\ ( h, k)\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}\\\\ -------------------------------\\\\ \begin{cases} h=-7\\ k=-3\\ p=8 \end{cases}\implies 4(8)[y-(-3)]=[x-(-7)]^2 \\\\\\ 32(y+3)=(x+7)^2\implies y+3=\cfrac{1}{32}(x+7)^2 \\\\\\ y=\cfrac{1}{32}(x+7)^2-3](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bparabola%20vertex%20form%20with%20focus%20point%20distance%7D%5C%5C%5C%5C%0A%5Cbegin%7Barray%7D%7Bllll%7D%0A4p%28x-%20h%29%3D%28y-%20k%29%5E2%0A%5C%5C%5C%5C%0A%5Cboxed%7B4p%28y-%20k%29%3D%28x-%20h%29%5E2%7D%0A%5Cend%7Barray%7D%0A%5Cqquad%20%0A%5Cbegin%7Barray%7D%7Bllll%7D%0Avertex%5C%20%28%20h%2C%20k%29%5C%5C%5C%5C%0A%20p%3D%5Ctextit%7Bdistance%20from%20vertex%20to%20%7D%5C%5C%0A%5Cqquad%20%5Ctextit%7B%20focus%20or%20directrix%7D%0A%5Cend%7Barray%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A%5Cbegin%7Bcases%7D%0Ah%3D-7%5C%5C%0Ak%3D-3%5C%5C%0Ap%3D8%0A%5Cend%7Bcases%7D%5Cimplies%204%288%29%5By-%28-3%29%5D%3D%5Bx-%28-7%29%5D%5E2%0A%5C%5C%5C%5C%5C%5C%0A32%28y%2B3%29%3D%28x%2B7%29%5E2%5Cimplies%20y%2B3%3D%5Ccfrac%7B1%7D%7B32%7D%28x%2B7%29%5E2%0A%5C%5C%5C%5C%5C%5C%0Ay%3D%5Ccfrac%7B1%7D%7B32%7D%28x%2B7%29%5E2-3)
keep in mind that the vertex is half-way between those two fellows, and the distance from the vertex to either one of them is "p" units, check the picture below.
with that focus point and that directrix, the half-way over the axis of symmetry will be -7, -3, that's where the vertex is at, and notice the distance "p", is 8 units.
since the parabola is opening upwards, "p" is positive 8.