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I.m not sure about  the log graph....but  I might go about it this way:

We can create a semi-log graph with the years being on the x axis (linear) and the distances  (as base 10 logs) are on the y axis

Let 1972  be  year 0

Let  2003  be year 20

Let 2009  be year  26

The distance from the Sun to the Earth  (the launch site of Pioneer 10) in 1972  is 140,000,000,000  m

Expressing this  as 10 raised to a power wehave  that   log 140,000,000,000  = 11.14

So ....the first point on the graph would be   (0, 11.14)

In 2003, Pioneer 10 was  11,933,236,224,000 m  from the Sun

So.....  log 11,933,236,224,000 ≈  13.08

So....the point (20, 13.08)  would be on this graph

In 2009  Pioneer 10  was  14.229.146,880,000 m  from the Sun

So....log 14,229146,880,000  ≈  13.15

So....the point  (26, 13.15)   would be on this graph

One last  thng.....   the distance from the Sun to Aldebaran  = 614,947,480,717,752,000 m

So....Log 614,947,480,717,752,000   = 17.78

This will not have a point associated  with it.....so.....I'll leave it to your discretion as how you want to show this

You would need to label the x axis as "Years Since 1972"

Label the y axis as " Base 10 Log Distances "

Here's the graph  (you would need to draw a curve through these points.....I don't know how to do that in Desmos) 

https://www.desmos.com/calculator/rsuu7k8yj6

That's about the best I can do with this   !!!

Thanks, EP!!!!!....nice explanation....

What does #Whoskeepingcount mean?????

Also this problem is a special case

x² - y² = (x - y) * (x + y)

Please do not advertise on this website...

Lol, NOW I know why "I am a robot" is added.

You should set a requirement of a minimum number of likes on your profile before the "I am a robot" disappears.

For your answers, all of your steps seem correct!

Does your teacher check for the exact answer? If you are a hundredth of a thousandth off, does your teacher consider it wrong?

Your answers and solution steps seem accurate, so you should be fine.

You can see what other people say though.

for the last part that says to make the logarithmic scale, how wuld i go forth in doing that ?

Not asking you to create the whole graph for me but as it goes for x and y axis, I can try my best to figure out the rest. Thank you

I just wanna make sure I know what im doing

If circumference is 14 pi       

Circumf = pi * d   so d = circumf / pi = 14     radius = 7

Circle with center at h,k and radius r     formula  is  (x-h)^2 + (y-k)^2  = r^2   (remember that?)

(x-10)^2 + (y-7)^2 = 7^2

Area = pi r^2      400pi /pi = radius^2 = 400      r=20

(x+4)^2 + (y+5)^2 = 400                                                          (edited typo)

the last two are similar and youo should try them yoursefl 

Everyting here makes sense! Way more sense then what I was thinking to solve this. I was making it way more complicated than it was.

I have addedd the three zero's as well!

Starting here at part 1, just did the calculations myself to double check and it all looks correct! I had solved this a bit differently but I beilive I was doing it all incorrect for this first step.

nvm its sqrt39

If the daylight factor in a house is 2%, then 2% of the 5000 lux will make it inside... or .02 as a decimal

So .02*5000 = 100 lux comes into the kitchen from outside.

Third Part

Aldebaran is about  65 light years from the Sun  (Wikipedia)

Number of meters in one light year ≈  9,460,730,472,580,800 m

So.....the number of approx. meters from the Sun to Aldebaran  ≈ 

9460730472580800  * 65 = 

614,947,480,717,752,000 m

Second part

Days between launch date and 1/1/2009  = 13455 days  = 1,162,512,000 seconds

So....the distance from the Sun to Pioneer 10  as of Jan 2009  ≈

Rate * Time  = Distance

12.24 km /s * 1,162,512,000 s  =   14.229.146,880 km

Jenny....just add three 0's to this to get the number of meters....!!!

Pioneer 10 launch date  = 3/2/1972

Days between this date   and 1/22/2003    =   11,284 days  = 270,816 hrs = 974,937,600 seconds

Assuming that the rate of  12.24 km/s   is the constant rate ( which is probably not constant)....the distance that Pioneer 10 was  from the Sun on 1/22/2003  =

Time  * Rate  = Distance

974, 937,600 s  * 12.24 km / s    ≈    11,933,236,224 km

So  multiplying this by 1000  for the number of meters we get

11,933,236,224,000  m

When I first solved it by adding i got a different value so I was confused, but now resolving it I did get the same value. I just wanted to make sure! Thanks(:

for the first part after I multiply the veloctiy to get the distance wouldnt I then add distance between Sun and Pioneer 10 to the 6 years the pioneer 10 has covered in distance?

No.....the first distance given  (  15,220,760,520,000 m )  =  the distance Voyager 2 was from the Earth as of 1/2/2009

We must add the distance from the Sun to the Earth ( 149,668,992,000 m)  to get the final distance that Voyager 2 was from the Sun as of this date = 15,370,429,512,000 m

I am going to have to make sure on the distances and all, but I can always jsut switch out the values since you have gave me the equations! Thank you.

I am curious as to why you included the part in bold

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

Do i subtract the dsitance from earth to sun on average by the 15,220,760,520,000 m to get the final answer that is underlined?

As it goes for question 4, since I am using AU units would I solve it differently ? and do 5.5786392/-0.413412695 ? and whatever that equals is the answer? or would I have to use the metre distance for that ?

Just finsihed up creating my own graph and as well jsut used the AU units, I felt it to be a bit easier that way. This is all making sense, thank you again for your help ! Theres still some more parts but I am going to go ahead and try doing it myself before I ask any questions.

As it goes for question 4, since I am using AU units would I solve it differently ? and do 5.5786392/-0.413412695 ? and wahteer that equals is the answer? or would I have to use the metre distance for that ?

See here :

https://www.desmos.com/calculator/vhar3yz5ku

Last one

This graph is definitely  strongly negative   !!!

Second one

This graph looks to have a moderate  positive correlation

First one

Plotting the training hours on x  and tthe accidents on y we have the following points

(0,8)  (12,0)  ( 2,5) (10, 1) (9, 3)  (5,2) (6,2) ( 4, 0)

See the graph below

The r value   =  -.7283   .....this indicates a strong negative correlation.....in other words.....the fewer the training hours, the more accidents

x^2 - 3x + 7  = 5 - x^2         add x^2 to both sides......subtract 5 from both sides

2x^2 - 3x + 2  =  0        (1)

The dicriminant   is

(-3)^2 - 4(2)(2)  =

9  -  16  =

-7

Thus.....if the discriminant is negative, we have no real solutions to   (1)....so.....there are no intersection points  as shown by the graph here :

https://www.desmos.com/calculator/1gyqxrcryq