Answer:
Step-by-step explanation:
Parameterize the ellipse as (acos∙,bsin∙). Take points P:=(acosp,bsinp) and Q:=(acosq,bsinq) on the ellipse, with midpoint M:=(P+Q)/2.
If |PQ|=2k, then
a2(cosp−cosq)2+b2(sinp−sinq)2=4k2
The coordinates of M are
xy==a2(cosp+cosq)b2(sinp+sinq)
The number 2852.1 is already rounded to the tenth since the tenth place is one place to the right of the decimal
8160 dived by 85 equals 96, 2880 divided by 30 equals 96, therefore she can make 96 necklaces
A)
To be similar triangles have to have equal angles
triangle ZDB'
1)angle Z=90 degrees
triangle B'CQ
1) angle C 90 degrees
angle A'B'Q=90
DB'Z+A'B'Q+CB'Q=180, straight angle
DB'Z+90+CB'Q=180
DB'Z+CB'Q=90
triangle ZDB'
DZB'+DB'Z=180-90=90
DB'Z+CB'Q=90
DZB'+DB'Z=90
DB'Z+CB'Q=DZB'+DB'Z
2)CB'Q=DZB' (these angles from two triangles ZDB' and B'CQ )
3)so,angles DB'Z and B'QC are going to be equal because of sum of three angles in triangles =180 degrees and 2 angles already equal.
so this triangles are similar by tree angles
b)
B'C:B'D=3:4
B'D:DZ=3:2
CQ-?
DC=AB=21
DC=B'C+B'D (3+4= 7 parts)
21/7=3
B'C=3*3=9
B'D=3*4=12
B'D:DZ=3:2
12:DZ=3:2
DZ=12*2/3=8
B'D:DZ=CQ:B'C
3:2=CQ:9
CQ=3*9/2=27/2
c)
BC=BQ+QC=B'Q+QC
BQ' can be found by pythagorean theorem
The real roots are 3 and -3.
Imaginary roots:-
x - (2 - i) = 0
x = 2 -i is one imaginary root and the other is 2 + i
