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OK!!!!

oh, duh! thanks! that helps a lot!

thank you!

happyredpanda! :)

Good job, CU   !!!

Lets do it the dumb way

25 - the digit's product is 10, so no.

50 - 0 , so no.

75 - 35, so no.

100 - 0 so no.

125, 150, and 175 and 200 are no because of our past pattern

225 - 20, so no.

250 - 0 so no.

275 - 70 so no.

300 - so no.

325 - 30, so no.

Notice the bolded, every 25 the product of its digits increase by 10.

So,

325 - 30

425 - 40

525 - 50.

525 is answer

Note that  ( .125)   =    125/1000  =   [25 * 5] / [ 200 * 5 ]  =  25 / 200   =   1 / 8 

So  we have  (1/8)^3  = ( 1/2^3)^3  = 1/2^9

1 / 2^9      =    [ remember that   1/ b^n  =  b^(-n)  ]....so..... 

1/2^9  =   2^(-9)

So  "a"  = -9

Here's my attempt  !!!

Mean  =   sum / number of integers  =  42/7  =    6

We would like to make the mode  (and median) as small as possible.....

Let the mode = 4  and the median   = 5

And we want the first  integer to be as small as possible  = 1

1  4  4  5  6  7 15

The mode  = 4

The median  = 5

The average = 6

I get 15  as the largest value.......???

Maybe CU  has a better answer    !!!

Take the elapsed time between its launch date and its contact-loss date. Then multiply that by its velocity, (make sure units match).

To calculate the position of pioneer 10 as of Janurary 2009, you take the elapsed time between the launch date and Janurary 2009, then take that and multiply it by the velocity to get the distance. (make sure units match)

The same methods can be applied for the other questions.

As for the logarithmic scale, I believe you either have learned to do that in class or in the past posts have learned it from other user's posts.

No prob!!!

I'm assuming that there may be a  more "mathematical" way to solve this......but....I never hit on it.....LOL!!!!

Yes, CPhill calculated it!

Why is the universe so big, but small at the same time.

I haven't learned this yet, so just curious, how would 2 be considered to be a complex number?

Lol! My math teacher made this    the last question on our algebra quiz last year! Brings back good memories of day dreaming in math class!

From Yahoo ( so....I can't vouch for the reliability)

At 0h UT (Greenwich time) on 2nd January 2009, Voyager 1 was 108.547628513 AU from the Sun. This is 16,238,494,019 km or 10,090,132,389 miles.

Also.....

From, Quora......as of Dec 1, 2017  .....Voyager 2 was traveling at about 34,390 mph= 55,345 km per hour.....it was slightly decelerating all the time 

Asuuming that the rate of deceleration is negligible.....there were  11,459 days   between its launch date of 8/20/77 and  01/02/2009

This is 275,016 hrs

So....from the Earth the (rough)  distance traveled (in meters)  =

275,016 hrs * (55,345 * 1000)m  =  15,220,760,520,000 m

And the distance from the Earth to the Sun (on average) is ≈  149,668,992,000 m

So.....on Jan 2,2009......Voyager 2 was about   15,370,429,512,000 m from the Sun

Ill check it out, thank you for yuor help!

I am so Boreedddd!!!! I want to type in all caps!!!

SChool is so eAsY!!

Why are teachers so evil, the lunch moniters are evil! I suffer from blasted eardrums during lunch, and enchaphlitis during class. My eyes are bloodshot because I like sleeping in to noon but school starts at 8:00
 

This is middle school for me, imagine highschool!
 

Even worse, imagine college....

The outlook on life isn't that great.

Sorry, my above answer is inaccurate, not necessarily incorrect.

Time is constantly moving forward, by 1 second each second

So the distance is ever changing if you used my method above.

Easier method:

Find when Voyager 1 and 2 were launched. (google)

Find speed (google)

Fond elapsed time between initial launch and 2009 januarary 2

Then Mulitply elapse time and speed, done.

what does #whoiskeepingcount mean?

Also, kminery19, try looking for same degrees to simplify your equations.

This method is inaccurate.

Find the current speed of voyager 1 and 2. (google)

Find the current distance right now (google)

Now, find the amount of time that has elapsed between Janurary 2, 2009 and right now.

Then do some maths

Hint: Take the elapsed time, multiply by the speed, and the subtract that from the current distance

You can use whatever scale you want.

The logarithmic graph I have done includes the stars as well as the plantets.

Yes you can change it to a number line.

Just push all the entitiies left so that they line up on the vertical axis.

Lol I know this isn't serious

hmmm....     (ignoring title)

its actually y^9 = x^6

Then....

y^3= x^2

As your typing your response, I was wondering, where can i findinformation about how far from the sun is voyager 1 and 2 as of january 2,2009 , in meters? I cant find any information on it. Thank you, sorry I know it isnt math related but I cant find any info.