Question

help

Answer 1

much easier
center is \((3,-5)\)

Answer 2

give me steps

Answer 3

standard form is \[\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\] for an ellipse with center \((h,k)\)
in this example \(h=3,k=-5\)

Answer 4

here \(a^2=64\) so \(a=8\) and \(b^2=100\) so \(b=10\)
because the larger number is under the \(y\) term the ellipse look like

Answer 5

do i put that..

Answer 6

since \(b=10\) the length of the major axis is \(20\) and since \(a=8\) the length of the minor axis is \(16\)

Answer 7

yes you can put that if you want
you need to to know what the graph looks like
i guess the last step is graphing

Answer 8

your graph should look like this
http://www.wolframalpha.com/input/?i=ellipse+(x-3)%5E2%2F64%2B(y%2B5)%5E2%2F100%3D1

Answer 9

label the center \((3,-5)\) and also label the vertices

Answer 10

tf

Answer 11

do just draw and egg and put 3,-5 in the middle

Answer 12

pretty much, yes

Answer 13

so no graph?

Answer 14

you mean graph paper?

Answer 15

um

Answer 16

here is what it looks like with a coordinate axis

Answer 17

yikes

Answer 18

don't you have any graphing tools?

Answer 19

scuse me. Its beautiful

Answer 20

whatever floats your boat

Answer 21

you could also label the vertices, but maybe you should just leave it alone

Answer 22

lol whats that

Answer 23

switch?

Answer 24

no what you have is good

Answer 25

i add that?

Answer 26

except the "since b =... " part should come under "it looks like this, not this"

Answer 27

what

Answer 28

what what?
because the larger number is under the y terms it looks is one part it should be together, not separated by "since a^2=64, a=8..."

Answer 29

next

Answer 30

center is \((6,0)\) just like the case of the ellipse

Answer 31

ok steps yall

Answer 32

put in order plz

Answer 33

like before
standard form of hyperbola with center \((h,k)\) is \[\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1\] in this case \(h=6,k=0\) so center is \((6,0)\)

Answer 34

thats all

Answer 35

for part a, yes

Answer 36

here \(a^2=25\) so \(a=5\)
length of the transverse axis is \(2\times 5=10\)
\(b^2=144\) so \(b=12\) length of conjugate axis is \(2\times 12=24\)

Answer 37

thats b?

Answer 38

asymptotes have slopes \[\pm\frac{a}{b}\] so slopes are \[\pm\frac{5}{12}\]

Answer 39

yes, that was b

Answer 40

and your graph should look something like this
http://www.wolframalpha.com/input/?i=hyperbola+y%5E2%2F25-(x-6)%5E2%2F144%3D1

Answer 41

or this if you want to include a coordinate axis
http://www.wolframalpha.com/input/?i=y%5E2%2F25-(x-6)%5E2%2F144%3D1

Answer 42

wanna do it cuz we all know mine will like like a kid on meth drew it

Answer 43

not that great but best i can do

Answer 44

ok but slopes should be \(\pm\frac{5}{12}\) not 512

Answer 45

lol

Answer 46

now it is my bed time

Answer 47

ok looks good?tho ?

Answer 48

yes, looks decent

Answer 49

Hey can u get on tmr same time, also sometimes my questioncove website doesnt work so plz go on peeranswers and see if im on there, if im not here

Answer 50

i will try

Answer 51

what is "same time"? like tonight? 9 ish for me

Answer 52

what time is it rn

Answer 53

11:30

Answer 54

its 8:30

Answer 55

in hawaii maybe

Answer 56

so 6 for me

Answer 57

ok
do you go to school during the day?

Answer 58

nop nothing this week

Answer 59

kk ttyl

Answer 60

what do you do all day?

Answer 61

k bye