Question

is this right?

Answer 1

nope
the first entry is correct but the second one is not.

Answer 2

yup

Answer 3

@Loser66

Answer 4

yup

Answer 5

Can you see the relationship among the elements?

Answer 6

yeah i think it is B or D but I'm not sure

Answer 7

ok, so test B

Answer 8

\(a_n=3*(4^{n+1})\)
Now, if I pick -12, that is \(a_2\) right?

Answer 9

yes

Answer 10

So, \(a_2= 3*(4^{2+1})=3*4^{3}=192\neq -12\)

Answer 11

Now, test D, you do

Answer 12

If you hate someone, do you want to hit him?

Answer 13

If someone scares you, do you want to hit him?

Answer 14

no

Answer 15

hahaha...

Answer 16

whats the point of these? can i please have help with my math

Answer 17

That is why you do not want to hit the problems scare you.
To me, if I don't know how to solve it, I hit it as many times as I can if it helps me to master the process.
I want you to practice.
Just test D by yourself

Answer 18

ok i tested D on my paper and it came out wrong again

Answer 19

If D doesn't work,
Test A, Test C until you get the answer
I am done

Answer 20

ok i tested all of them and got C!

Answer 21

Bingo.
You will go to standford!!

Answer 22

yay- thank you!!!

Answer 23

@Loser66

Answer 24

As usual, tell me your opinion first

Answer 25

I believe it is D @Loser66

Answer 26

why?

Answer 27

Substitute a = 25 and b = 15 in the ellipse equation.
x2/(25)2 + y2/(15)2 = 1
x2/625 + y2/225 = 1

Answer 28

When they ask you about the vertices and foci at the same time, they meant the major vertices.

Answer 29

ok

Answer 30

So that, if foci are on the y -axis, then the vertices MUST on the y-axis also.
for D) foci are \((0,\pm 15)\) so, \(c=\pm 15\), and \(c=\sqrt{a^2+b^2}=\sqrt{625+225}=29.15\neq 15\)

Answer 31

I give you my trick

Answer 32

the number under y^2 is 625, right? and 625 >225, right?
so, the foci are on the y axis. ok?

Answer 33

So, the foci is under the form of (0, something)

Answer 34

hahaha...such a good future Standford student!!!

Answer 35

you got my medal for that correct answer

Answer 36

hey, greedy guy!! I am busy now.
just 1 more!!

Answer 37

shoot

Answer 38

for cos, it is ALWAYS symmetric about x -axis.

Answer 39

and you need know the rules:
About x-axis: both \((r, \theta), (r,-\theta)\) are on the graph.

Let's test \(r= 5 cos (5\theta) \), if \(\theta =-\theta\), we have \(cos (\theta)=cos (-\theta)\)
Hence \(r = cos(-5\theta)\)

So, \((r,-\theta)\) is also on the graph.

About y-axis:
(-r, -theta) must be on the graph. now cos (-5theta) = cos 5theta = r
and \(r\neq -r\), so it is not about y-axis.

About origin
(-r, theta) Must be on the graph.

\(-r = - 5 cos (5 theta)\neq r\)

Hence not about origin

Conclusion: just about x-axis.