hectictar

Thank you!!

2)  This one is tricky but here's my attempt...

I started by making this list:

By now we can notice something...

Between  4  and  9 ,  there are  4  numbers and  4/2  =  2

Between  9  and  16  there are  6  numbers and  6/2  =  3

So the number of 3's   =  1 + 2 + 3   =   6

We can backtrack the previous steps to get...

the number of 3's   =   \(1+\frac{4}{2}+\frac62\ =\ 1+\frac{9-4-1}{2}+\frac{16-9-1}{2}\ =\ 1+\frac{3^2-(3-1)^2-1}{2}+\frac{(3+1)^2-3^2-1}{2}\)

So in general,

the number of  n's   =   \(1+\frac{n^2-(n-1)^2-1}{2}+\frac{(n+1)^2-n^2-1}{2}\)     which simplifies to...

the number of n's   =   2n

This agrees with the list from earlier!!!

Except the number of  45's  is only  27  because the list ends at  b(2007)

\(\sum\limits_{p=1}^{2007}b(p)\ =\ b(1)+b(2)+b(3)+b(4)+b(5)+b(6)+b(7)+\dots+b(2006)+b(2007)\\~\\ \sum\limits_{p=1}^{2007}b(p)\ =\ 1+1+2+2+2+2+3+\dots+45+45\\~\\ \sum\limits_{p=1}^{2007}b(p)\ =\ 2(1)+4(2)+6(3)+\dots+88(44)+27(45)\\~\\ \sum\limits_{p=1}^{2007}b(p)\ =\ \sum\limits_{k=1}^{44}(2k)(k)\quad+\ 27(45)\\~\\ \sum\limits_{p=1}^{2007}b(p)\ =\ \sum\limits_{k=1}^{44}(2k)(k)\quad+\ 27(45)\\~\\ \sum\limits_{p=1}^{2007}b(p)\ =\ 2\sum\limits_{k=1}^{44}k^2\quad+\ 27(45)\\~\\ \sum\limits_{p=1}^{2007}b(p)\ =\ 2(\frac{44(44+1)(2(44)+1)}{6})\quad+\ 27(45)\\~\\ \sum\limits_{p=1}^{2007}b(p)\ =\ 58740\quad+\ 1215\\~\\ \sum\limits_{p=1}^{2007}b(p)\ =\ 59955 \)

P.S. There definitely might be a better way to do it...

1)  ( Assuming you meant compute f(5,5) )

k  =  (6x + 12)(x - 8)

                                          Multiply out the right side of the equation.

k  =  6x² - 48x + 12x - 96

                                          Combine like terms.

k  =  6x² - 36x - 96

In an equation of the form   k  =  ax² + bx + c   with   a > 0,

the least possible value of   k   occurs at   \(x\ =\ \text-\frac{b}{2a}\)

So...

In an equation of the form   k  =  6x² - 36x - 96   with   6 > 0,

the least possible value of   k   occurs at   \(x\ =\ \text-\frac{(\text-36)}{2(6)}\ =\ \frac{36}{12}\ =\ 3\)

When  x = 3,     k   =   6(3)² - 36(3) - 96   =   6(9) - 36(3) - 96   =   54 - 108 - 96   =   -150

Using the rule  tan(angle)  =  opposite / adjacent ,  we can make two equations:

\(\tan(61^\circ)\ =\ \frac{h}{x}\\~\\ x\tan(61^\circ)\ =\ h\\~\\ x\ =\ \frac{h}{\tan(61^\circ)}\)

...and...

\(\tan(32^\circ)\ =\ \frac{h}{600+x}\\~\\ (600+x)\tan(32^\circ)\ =\ h\\~\\ 600+x\ =\ \frac{h}{\tan(32^\circ)}\\~\\ x\ =\ \frac{h}{\tan(32^\circ)}-600\)

Now we can equate both expressions of  x  and solve for  h:

\(\frac{h}{\tan(61^\circ)}\ =\ \frac{h}{\tan(32^\circ)}-600\\~\\ \frac{h}{\tan(61^\circ)}-\frac{h}{\tan(32^\circ)}\ =\ -600\\~\\ h(\frac{1}{\tan(61^\circ)}-\frac{1}{\tan(32^\circ)})\ =\ -600\\~\\ h\ =\ -600\div(\frac{1}{\tan(61^\circ)}-\frac{1}{\tan(32^\circ)})\\~\\ h\ \approx\ 573.5998\\~\\ h\ \approx\ 574\)

Let the y-intercept be the point  (0, a)  where  1 ≤ a ≤ 9

So the slope   =   \(\frac{\text{rise}}{\text{run}}\ =\ \frac{365-a}{4-0}\ =\ \frac{365-a}{4}\)

To make the slope as small as possible, we want to make the fraction  \(\frac{365-a}{4}\)  as small as possible.

To make the fraction as small as possible, we want to make the numerator as small as possible (we can't change the denominator).

And to make the numerator as small as possible, we want to subtract as much as we can from  365 .

To make  365 - a  as small as possible, we want to make  a  as large as possible.

The largest possible value of  a  is  9 ,  so the smallest possible value of the numerator is  365 - 9

And the smallest possible slope  =  \(\frac{365-9}{4}\ =\ \frac{356}{4}\ =\ 89\)

It's kind of hard to see the difference but here is a graph where you can see what changing the value of  a  does to the slope:

Using the information that the line passes through the points  (0, 0)  and  (-4, 3) ,  we can find the slope.

slope  =  \(\frac{\text{rise}}{\text{run}}\ =\ \frac{y_2-y_1}{x_2-x_1}\ =\ \frac{3-0}{-4-0}\ =\ -\frac34\)

Using the point  (0, 0)  and the slope  -\(\frac34\) ,  the equation of the line in point-slope form is

y - 0  =  -\(\frac34\)(x - 0)

Simplify both sides of the equation to get

y  =  -\(\frac34\)x

a + ab²  =  40b

a - ab²  =  -32b

The purple values are equal and the blue values are equal.  purple + blue  =  purple + blue

(a + ab²) + (a - ab²)  =  40b + -32b

(a + ab²) + (a - ab²)  =  40b + -32b

2a  =  8b

\(\frac14\)a  =  b

Now we can substitute this value for  b  into one of the original equations.

a + ab²  =  40b

                                    Substitute   \(\frac14\)a   in for   b

a + a(\(\frac14\)a)²  =  40(\(\frac14\)a)

                                    Simplify both sides of the equation.

a + \(\frac{1}{16}\)a³   =   10a

                                    Multiply through by  16

16a + a³  =  160a

                                    Subtract  16a  and subtract  a³  from both sides

0  =  144a - a³

                                    Factor  a  out of both terms on the right side

0  =  a( 144 - a² )

                                           Factor   144 - a²   as a difference of squares

0  =  a( 12 - a )( 12 + a )

                                           Set each factor equal to  0  and solve for  a