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Ok, oops, I did not notice it was 9x^2007/x^2008 and 8x^2009/6x2010 .... Another careless mistake...

And also, I did it in a different way. I sepreated the numbers and the x's. Then I simplify the fractions so I didn't have to divide. 

Thanks so much for checking my answer!

Thanks RainbowPanda  

f(2) = 4

2 + (m - 1)*3 =3 + (n -1)*7, solve for m, n

3 m + 3 = 7 n

m =160 and n =69

2 + (160 -1)*3 =479 - the 160th term in the first AP

3 + (69 - 1)*7  =479 - the 69th term in the second AP.

Note: The terms are both LESS  than 500. However, 500 itself is shared between the two APs.

Herons Formula, when so applied, is something completely new to me.
Thank you, Mathhhemathh!

  !

thank you that helped alot 

3 r - 19 = k/(2 r + 1) ............(1)

6 r - 27 = k/(r - 5) ...............(2)

r =7 and k =30

I have the same queston

Tysm for all of your help! I'm guessing you want me to do the cable cars avg rate of change for 5a

First of all...the cable car's  avg rate of change is constant....it is  straight line

It changes  by 50 ft per second  over all the given intervals

Pat's rate of change  from 0 to 2

[ 125 - 0 ] / [ 2 - 0 ]  =  125 / 2  =  62.5 ft /min

From 2 to 4

[ 200 - 125 ] / [ 4 - 2 ] =  75/2  = 37.5 ft / min

From 4 to 5

[ 225 - 200] / { 5 - 4 ]  =  25 /1  =  25 ft / min

Looks like Pat is running out of gas at the end while the cable car maintains the same steady pace.....LOL!!!

b Pat's rate of change is not constant.....the cable car's rate of change is

c Pat's average rate over the ntire trip  =  [ 250 - 0 ] / [ 6 - 0 ] = 250 / 6 ≈ 41.66 ft / min

As before....the cable car's rate  =  50ft / min

um I think I was doing the wrong problems XD 

Thank you so much!

3   Avg  rate of Temp Change  =  [ Ending Temp  - Beginning Temp ] / [ Number of elapsed hours]

    [ 52  - 72 ]  / 3  =  -20/3  ≈  -6.66 degrees per hour

4   [ Value at the end - Value at the beginning ] / [ Number of years elapsed ]

   [ 20050  - 12500 ] / 19  ≈ $397.37  per year

3. 8-4/1-(-2)=4/3

4. 1450-1140/4-1=310/3=$103.33

$103 per month

Am I right? (I tried lol)

I really appreciate all of your help!

OK...glad to  help where I can....

Thank you! I honestly only wanted the formula, but thanks haha

I wanted to make sure I was doing it correctly ^-^

a. Similar to the last one, we have

Avg rate of change   = [Ending height - Beginning height ] / [ Ending time - Beginning time ]

First 2.5 seconds

[25 - 0 ] / [ 2.5 - 0]  = 25 / 2.5   = 10 ft/sec

2.5 to 5

[ 0 - 25 ] / [ 5 - 2.5 ] =  -25 / 2.5  =  -10 ft/sec

0 to 1

[ 15 - 0 ] / [ 1 - 0 ]  =  15 ft/sec

1 to 2

[ 24 - 15 ] / [ 2 - 1 ] =  9/1  = 9 ft/sec

b. Definitely not.....the graph is a parabola.....the rates of change on a parabola are not constant

c. Avg rate of change  over entire 5 seconds

[  0  - 0 ]  [ 5 - 0 ] = 0/ 5   =  0 ft/sec

Alternatively, you could do it algebraically.  Equate the two expressions for y:

x²+k = 1-kx

x² + kx + k-1 = 0.  (1)

The discriminant  of this quadratic is k²-4*1*(k-1) or k²-4k+4.  This is zero when k = 2.

Since there is only a single solution to equation (1) when the discriminant is zero, k = 2 doesn’t allow the original two equations to intersect in more than one point.

No prob , RP  !!!

Nice approach, Mathehemathh    !!!!

Thank you so much for helping, I totally understand this now :D

Rate of change  = [ final distance -  beginning distance ] / [ final time - beginning time ]

a. Rate of change over the whole trip  =  Miles / Hours  = 350 / 8  = 43.75 mph

b.  Between 4th and 5th hour  =  [300 - 225 ] / [ 5 - 4 ] = 75 / 1  = 75 mph

     Between the  5th and 6th hour  = [300 - 300] / [ 6 - 5]  = 0 / 1  = 0 mph

     Between the   6th and 8th hour = [ 350 - 300 ] / [ 8 - 6 ]  = 50/2  = 25mph

c.  No....if they were....the line would have a constant slope....but it doesn't

d. No...for the same reason given in (c)

The graph above might help (even though it is a little busy!).