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Are you happy with these answers ?   

2 is the best answer   :)

Yes how big are the rectangles and how big is the grid?

Nothing because there is no x or y 

Hi Tetration

What does double factorial mean ?

4.9 t^2+25 sin(30)

$$\\=4.9 t^2+25\times 0.5 \\
=4.9 t^2+12.5 \\$$

i entered that as   log(6,10)

$${{log}}_{{\mathtt{10}}}{\left({\mathtt{6}}\right)} = {\mathtt{0.778\: \!151\: \!250\: \!383\: \!643\: \!5}}$$               

Continued here

Yes i am happy with my answer I think.   

I have played with the graph some more as well and I am totally happy.

I did this the same as you and found....as you did...that whatever values were put into the original function to produce "6" - namely, ±√3 and 2 -  were outside the prescribed domain of the original function...!!!!

Maybe we have "clunker" here ?????

If x ≠ 0, what is the solution to the following equation? (1 - x)/x + 2 = 7/x

Getting a common denominator on the left, we have

(1 - x)/x + 2x/x  = 7/x    subtract 2x/x from both sides

(1 -x)/ x  = 7/x - 2x/x

(1-x )/x  = (7 - 2x) /x     cross-multiply

x(1 - x)  = x(7 - 2x)    simplify

x - x^2  = 7x - 2 x^2  rearrange

x^2 - 6x = 0  factor

x(x -6) = 0     set both factors = 0...so x = 0 or  x = 6

And since x can't = 0,   then x = 6 

                      -

I am sticking with my first answer - I think that the algebra is correct

My original graph had the wrong coefficients so it was incorrect ***

This graph is correct and it shows that there are no values for      $$f^{-1}(6)$$

2-1=1 apple left

Melody...I don't think it's going to be very easy to "manually" isolate  "x" in the function -x^3 -2x^2 + 3x so that an inverse an be found  ..!!!!

Here's what WolfamAlpha generated as the inverse.....it looks pretty  "nasty".... (to say the least)

I think Alan answered this one.....but he did it graphialy, as I recall....

$$\int\frac{1}{x}dx$$

If x is positive we both agree that the answer is ln(x) +c

 If x is negative the answer should be    -ln(-x)+c    OR     -ln|x|+c       shouldn't it ?

9+10=

21

lol

What's the question?

Starting with  -10x² + 11x  =  -8,  we can get  -10x² + 11x + 8  =  0  that can be represented by the function:

y  =  -10x² + 11x + 8.  This give a graph of a parabola which has an infinite number of pairs of x-values that

have the same y-value (on the same horizontal line) from ground level to its maximum.

I'm not understanding what you mean by the "same horizontal plane". 

The general integral of  1/x  is  ln|x|  because ln() must contain a positive value.

If we know that x is positive, we don't need the absolute value bars.

Hi Geno,

why have you used     ln|u|     instead of   ln(u)   ?

∫[ x/(2 + x) ] dx

Let  u  =  2 + x   --->  x  =  u - 2         and        du  =  dx

∫[ x/(2 + x) ] dx   --->   ∫[ (u - 2)/u ] du   --->   ∫[ u/u - 2/u ]du   --->   ∫1 du  -  2∫1/u du

   --->   u  - 2 ln|u| + C   --->   2 + x - 2 ln| 2 + x | + C

I think that this question was answered in the last couple of days already. (By someone else)

BUT

y=3-x  This is a line with gradiene = -1 so when it is reflected across y=x it stays the same

So its inverse is y=3-x

$$f(0)=3\qquad so \qquad f^{-1}(0)=3$$    

$$\\y=-x^3+2x^2+3x\\
inverse\;\; function\\
x=-y^3+2y^2+3y\\
when\;\; x=6\\
6=-y^3+2y^2+3y\\
0=-y^3+2y^2+3y-6\\
0=-y^2(y-2)+3(y-2)\\
0=(-y^2+3)(y-2)\\
3-y^2=0\qquad or \qquad y-2=0\\
y^2=3\qquad \qquad or \qquad y=2\\
y=\pm\sqrt{3}\qquad \qquad or \qquad y=2\\
y=\pm1.732 \qquad \qquad or \qquad y=2\\
but\;\; y>3 \\
$So no solutions to $ f^{-1}(6)$$

NO SOLUTION.

I have replace the graph the original one had the wrong coefficients and caused me great confusion.