Question

how many three digit number can be formed from 1,2,3,4 and 5 if repetition is allowed?

Answer 1

sorry brother sister hard for me to understand english + math

Answer 2

i cant speak english (i from indonesia), but i love MATH :p

Answer 3

your good me nosebleed sometime hahaha

Answer 4

5^3

Answer 5

rad, can you help with parametric equations

Answer 6

thanks eddie

Answer 7

right 5^3 = 125
what ur question eddie

Answer 8

i want to convert a parametric equation to a polar

Answer 9

for example ?

Answer 10

, x = 4sin t + 2sin(2t) , y = 4cos (t) + 2cos(2t)

Answer 11

i tried doing x^2 + y^2 , but it wouldn't eliminate t.
the directions say change the parametric equation to polar equation

Answer 12

how many four digits number can be formed from 1,2,3,4 and 5 if no repetition is allowed?

Answer 13

5 choices for first number, then 4 choices, then 3 choices, then 2 choices
5x4x3x2

Answer 14

thanks eddie

Answer 15

coordinat polar like (r, alpha) ? with r=sqrt(x^2+y^2)
alpha = arctan (y/x), right eddie ?

Answer 16

, yes

Answer 17

so my idea was to change parametric to cartesian, then cartesian to polar. but it didnt work

Answer 18

if there are three routes from manila to pampanga and also three from pampanga to pangasinan.How many different routes may be taken if one wants to travel from manila to pangasinan?

Answer 19

3x3

Answer 20

you finish my assigment eddie 3 more to go yehey =))))

Answer 21

wahahaha

Answer 22

x^2 = [4sin t + 2sin(2t)]^2 = 16sin^2 t + 16sintsin2t + 4sin^2 2t
y^2 = [4cos (t) + 2cos(2t)]^2 = 16cos^2 t + 16costcos2t + 4cos^2 2t
x^2+y^2 = 16(sin^2 t + cos^2 t) + 16(sintsin2t+costcos2t) + 4(sin^2 2t + cos^2 2t) = 16(1) + 16cos(t-2t) + 4(1) = 16 + 16cost + 4 = 20 + 16cost

Answer 23

I got x^2 + y^2 = 20 + 16cos(t)

Answer 24

, but now we still have that t , we want to eliminate that

Answer 25

yes, twine mine

Answer 26

twine mine
?

Answer 27

nopes... yours like m :)

Answer 28

ok :)

Answer 29

where go from here

Answer 30

actually, r be scalar... but idk why there is t there :)

Answer 31

yeah im stumped

Answer 32

r^2 = x^2 + y^2
r = sqrt(x^2 + y^2)
r = sqrt(20+16cos16t) = sqrt(4(5+4cos16t)) = 2sqrt(5+4cos16t)
just simplify it :)

Answer 33

topic:permutation

A building has 5 entrances and 4 exit. In how many ways can you go in and out of the building?

Answer 34

wait, how did you get sqrt ( 20 + 16 cos 16 ?

Answer 35

r = sqrt( 20 + 16 cos (theta)) and that is wrong, i just checked the graph

Answer 36

pythagorean theorem :
r^2 = x^2 + y^2
r = sqrt(x^2 + y^2)

Answer 37

r = sqrt( 20 + 16 cos (theta)) is not the same as the graph of
x = 4sin t + 2sin(2t) , y = 4cos (t) + 2cos(2t)

Answer 38

sorry there is typo above :)
r = sqrt(20+16cost) = sqrt(4(5+4cost)) = 2sqrt(5+4cost)

Answer 39

ok, that is the wrong graph, the polar is not the same as the parametric

Answer 40

I think i got it

Answer 41

from the second eqution we can solve for cos (t)

Answer 42

y = 4cos (t) + 2cos(2t)
now change cos (2t) = 2 cos^(t) - 1

Answer 43

so y =4cos(t) + 4 cos^2(t) -2

Answer 44

then let it equal to zero

Answer 45

4cos^2(t) + 4cos (t) -2 - y = 0 , solve for cos t

Answer 46

I got it, but its pretty long solution, message me if youre interested

Answer 47

yea, its very interesting... great

Answer 48

want to see it ?

Answer 49

i think u can write it here :)

Answer 50

so to solve cos t from 4cos^2(t) + 4cos (t) -2 - y = 0 , that

Answer 51

yes

Answer 52

so far we have
x^2 + y^2 = 20 + 16 cos(t)
and from the second equation we have y = 4cos (t) + 2cos(2t) ,
substitute cos(2t) = 2cos^2(t) - 1

Answer 53

now from this
4cos^2(t) + 4cos (t) -2 - y = 0
use quadratic formula, let u = cos t
4u^2 + 4u -2 - y = 0, treat y as a constant , u is the variable
u = [-4 + - sqrt( 4^2 - 4*4*[-2-y] ) ] / (2*4)

Answer 54

ahhh, yeah i think i have conected ...
quadratic formula, i was forget it

Answer 55

now simplify that

Answer 56

u = -4/ 8 + - sqrt( 16 + 16(2+y)) / 8

Answer 57

u = -1/2 + - 4 sqrt(1 + 2+ y ) / 8
u = -1/2 + - 4 sqrt(3+y)/8

Answer 58

so cos t = -1/2 + - sqrt(3+y)/2

Answer 59

so far, i was so good :)

Answer 60

now lets plug this into x^2 + y^2 = 20 + 16 cos (t)

Answer 61

x^2 + y^2 = 20 + 16 [-1/2 + - sqrt(3+y)/2 ]

Answer 62

right, n then just take sqrt ?, im not sure

Answer 63

, then distribute

Answer 64

x^2 + y^2 = 20 + 16*-1/2 +- 16/2 sqrt(3+y)

Answer 65

x^2 + y^2 = 20 - 8 + - 8 sqrt(3+y)

Answer 66

x^2 + y^2 = 12 + - 8sqrt(3+y) , now plug in x^2 + y^2 = r^2, and y = r sin theta

Answer 67

r^2 = 12 + - 8 sqrt(3 + r sin theta) , and thats a polar equation

Answer 68

and i graphed that on wolfram, and it works. notice that this is is not a function r = f(theta),
unless you went real crazy and kept going,
r^2 - 12 = + - 8 sqrt(3 + r sin theta) , square both sides

Answer 69

make sense for m now, eddie..
what grade ur class, eddie

Answer 70

im in college

Answer 71

sophomore

Answer 72

so honestly i dont think the teacher gave this problem, or if he did , this is very difficult. its a rotated cardiod

Answer 73

i bet some rotation transformation might make it easier

Answer 74

yea, therefore i have to give medal for u today, eddie :)
great, that's amazing

Answer 75

so we learned a lot today
i think , never give up . but some problems are intractable , or you need some other method

Answer 76

new question

Answer 77

Write the polar equation r = cos(4*theta) in the parametric form x = x(t), y = y(t)

Answer 78

i was too old to more studying again...

Answer 79

we must study further

Answer 80

where are you from rad?

Answer 81

i from indonesia

Answer 82

sorry, if my english soo bad :)

Answer 83

thats ok , so lets multiply both sides by r
r = cos (4theta)
r^2 = r cos (4theta), but we have a problem, r cos theta = x , not r cos (4theta)

Answer 84

sorry, eddie... i want going to my student's home
there is a private lesson now, next time we can discuss again

Answer 85

nice to know u eddie

Answer 86

try post it in new posting, the other members maybe can help u