CPhill

Each orange takes up 4/3 as much volume as each pear

Thus....each orange can be considered as 4/3 of a pear

Therefore....if there already 28 pears in the basket, it could hold 12 more pears

But.......we could only have 9 more oranges in the basket because 9 (4/3)pears = 12 pears

I assume you have this :

√[20 q¹⁰ ]     ......if so   ..... this equates to....

(20 q¹⁰) ^1/2   =

√20  * q⁵  =

√[4 * 5] * q⁵  =

√4 * √5 * q⁵  =

2√5 q⁵  

y = (ln(x) )^x      take the ln of both sides

ln(y)  =  ln [ln(x)]^x

ln (y)  = x ln [ln (x) ]     now.....differentiate both sides

y' / y   = ln [ln (x) ]   +  x [  (1/x) / ln(x) ]    simplify

y' / y = ln [ ln (x) ] + 1/ [ln (x) ]    multiply both sides by y

y' = y [ ln [ln (x)  + 1/ [ln (x) ]      and substitute  [  ln(x) ]^x   for y

y'  = [ln (x)]^x [ ln [ln(x)] +  1 / ln (x) ]

Ordering of functions  from  slowest growing to fastest growing  as x  →  infinity

 e^(x/2)        e^x       [ln (x)]^x         x^x

It's not a solution to (b) or (d)......

Nice, fiora....!!!!

The linear function we are dealing with is :

y = 8x + 35

So....just put the first value for each ordered pair into the equation for x and see if it produces the second value of the ordered pair....if it does, that point is on the graph...

For instance....for the point (7, 91) we have

y = 8(7) + 35  =  56 + 35  =   91.......so this point is on the graph......

Let's suppose that it could be....then

5√9 = 5*3  = 15.........and this must mean that.....

15 = √45       square both sides

225  = 45     and this clearly isn't true

Distance =

√ [ ( 10 - 4)^2 + ( 13 - 5)^2 ]    = √ [ 6^2 + 8^2 ]  = √ [36 + 64]  = √100  = 10 units

The set up is:

20 + .50M   = 10 + .80M     subtract .50M, 10 from both sides

10  = .30M     divide both sides by .30

10 / .30  = M   =  about 33 + 1/3 miles driven equalize the cost

[Jana used  " -.50M"  in the original set up instead of  " +.50M"   ]