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a1=3  , an = a_n-1  * 1/6

y =  -2x^3 +6

y - 6 = -2x^3

[ 6 - y ] / 2 = x^3

∛ ( [ 6 - y ] / 2 )  = x

∛ ( [ 6 - x] / 2) = y = the inverse

Original into inverse

∛ ( [ 6 - (    -2x^3 +  6     )   ] / 2)

∛ ( [ 2x^3 ] / 2 ) =

∛ x^3 = x

Inverse into original

-2(  ∛ ( [ 6 - x] / 2)   )^3 +  6  =

-  ∛ [ 6 - x]^3  + 6  =

- 6 + x + 6    =    x

Nevermind guys. That's wrong too. 

It is impossible for Pat to select both 3 oranges and 6 apples, so we can calculate the probabilities of these mutually exclusive cases separately and then add to get our final answer. The probability that 3 particular pieces of fruit will be oranges and the rest will not be is given by (1/3)^3(2/3)^5=32/6561, and there are C(8,3)=56 ways of selecting three pieces of fruit to be the oranges so the probability that 3 will be oranges is 56*(32/6561)=1792/6561. Similarly, the probability that 6 particular pieces of fruit will be apples and the other two won't be is given by (1/3)^6*(2/3)^2=4/6561 and there are C(8,6)=28 ways of selecting which ones will be the apples, so multiplying again gives us a probability of 28*(4/6561)=112/6561. Adding those two probabilities give us our final answer: 1792/6561+112/6561=1904/6561. 

The volume of the can is the same whether it is full or empty....

the volume of the paint is

pi r^2 h     and h = .3 (30)

pi 5^2 .3(30 ) = 706.5 cm^3

3)  square root of (3x+7)

y = √[3x + 7]        square both sides

y^2 = 3x + 7

y^2 - 7   =  3x

[ y^2 - 7 ] /3 = x

[ x^2 - 7 ] / 3 = y  = the inverse

Put the inverse into the original

√ (  3 [ [ x^2 - 7 ] / 3  + 7 )  =

√  [ x^2 - 7 + 7 ] =

√x^2   = x

Put the original into the inverse

[ ( √[3x + 7]  ) ^2 - 7 ] / 3

[ 3x + 7 - 7 ) /3

3x /3 =   x

Volume of the cylinder = pi r^2 h

  pi 4^2  (3) = 48pi = 150.7 ft^3    (using pi = 3.14)

lim      [cos(3x) - 1] 

x→0  __________

            sin (4x)    

Apply L'Hopital's Rule and we have

lim      [-3sin (3x) - 0]              -3(0) 

x→0  ____________    =      _______   =     0     

              4cos (4x)                    4 (1)

I'm absolutely clueless on what to do or where to start

2.

x^2 - x - 2  =

(x  - 2) ( x + 1)

We have zeros at x  = 2  and x = -1

The only graph where this is true is the one at the top left

Big Brother, where does the problem say to multiply anything by –5 ? 

We aren't looking for the inverse function f^-1     we are speaking of the derivative of the function f '       

hm this is hard 

if a function f(x) is mapping x to y, then the inverse function f(x) maps y back to x

$y=\frac{15}{1+4e^{-0.2x}}$

interchange the variables x and y 

$x=\frac{15}{1+4e^{-0.2y}}$     solve for y 

$y=-5\ln \left(\frac{-x+15}{4x}\right)$

$-5\ln \left(\frac{-x+15}{4x}\right)$

find the domain of each inverse function 

domain of    ​\(​​​​​​v​\begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:

the domain of a function is the set of input or argument values for which the function is real and define

find positive values for log  \(0

$\log _af\left(x\right)\quad \Rightarrow \quad \:f\left(x\right)>0$

$\frac{-x+15}{4x}>0$

multiply both sides by 4 

$\frac{4\left(-x+15\right)}{4x}>0\cdot \:4$

simplify: 

$\frac{-x+15}{x}>0$

OMG i cant do this anymore
 

srry , thats all i know

1. 

x^2 + 5x - 6             ( x + 6)( x - 1)

__________  =       ___________

x^2 - x - 6                (x - 3) ( x + 2)

We will have zeros at x = -6 and x = 1

We will have a y intercept of (0, 1)

We will have vertical asymptotes at  x = 3 and x = -2

The graph on the bottom left is correct....

domain ​\(\begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:

Lol, well I hope I did my best to help :)

2) 

y =  (-3x+1)^2 - 2

y + 2 =  (-3x + 1)^2       take the root

sqrt ( y + 2) = -3x + 1

sqrt (y + 2) - 1 =  -3x

[ 1 - sqrt (y + 2) ] / 3  =  x           swap   x and y

[ 1 - sqrt (x + 2) ] / 3 =  y         this is the inverse

Put the original into this 

[ 1 - sqrt ( [  (-3x+1)^2 - 2 ] + 2 ] / 3  =

[ 1 - sqrt [ -3x  +1 }^2  ] / 3

[1 - [ -3x + 1 ] ] / 3

3x /3     =  x

Put the inverse into the original

(-3 [ 1 - sqrt (x + 2) ] / 3 ]  +1)^2 - 2     = 

( - [ 1 - sqrt ( x + 2) ]/3  ] + 1 )^2  - 2    =

( sqrt (x + 2) )^2 - 2  =  

x + 2 - 2   =   x

Step 1: Simplify both sides of the equation.

−5(x−7)=49

(−5)(x)+(−5)(−7)=49(Distribute)

−5x+35=49

Step 2: Subtract 35 from both sides.

−5x+35−35=49−35

−5x=14

 Step 3: Divide both sides by -5.

$\frac{-5x}{-5}=\frac{14}{-5}$

x= -14/5

answer is x=-14/5     

Here is the funtion and the graph of the derivative  and the line  x = ln4/(.2)

so each person get 71 cookies  

I don't think there is any distributive property involved in PEMDAS.

Say, in this problem here;  4 x 8 (2+6)

Well, PEMDAS says you have to get the parenthesis taken care of first, even if you immediately think you have to distribute the 8 through the parenthesis. :)

I hope this explanation helps :)

$\text{you need to find out where the 2nd derivative of }f \text{ is positive}\\ \text{It's the 3rd choice}$

Probably... he seems to be typing some massive MLA essay ;)

ElectricPavlov, what result did you get with your derivative calc?