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Just put in 77 for y ...then solve for 't ' using the Quadratic formula

77 = - 6t² + 43t   

-6t² +43t -77 = 0 

    a = -6     b = 43    c = -77

\(t = {-b \pm \sqrt{b^2-4ac} \over 2a}\)          throw out any negative 't' values

I hope you actually looked at the method of solution and understand !

You are welcome .....

The top is  equivalent to  482/99

the bottom is                  392/99

(482-392) / 99 = ...........            (it's    .909090909.......)

Thank you!!

at   point   0,5

       5 = a(0²) + b (0)   + c

       5 = c

Now sub in the point  1,10

   10 = a(1^2) + b(1) + 5

   10  = a + b + 5

     5 = a + b

Now  put in the point 2,19

19 = a (2^2) + b(2) + 5

19 = 4a + 2b + 5

14 = 4a+ 2b

Now you have two (RED) equations and two unknowns (a and b)   solve for a and b

a+b = 5

4a+ 2b = 14        I'll solve by 'elimination'    

                              ....   multiply the first equation by -2   then add to the second equation to get :

2a = 4

a = 2               and since   a+b = 5    b = 3       and  c = 5

See your original post for further help.....

a+b =8

    b = 8-a

a+c = -5

    c = -5 - a

b+c = -3

 substitute  

   8-a   +  -5 -a  = -3

   3 - 2a = -3

      -2a = -6

         a  = 3           you  should be able to continue from here.......

There's no actual number (not that I know of).

Expression if that's what it's asking: 12 * n.

Well, as you can see, the function is a line

y = mx + b    is  slope intercept form of a line   m = slope  b = y axis intercept (when x = 0)

Find two 'convenient' points like   -3,2    and   0,-4

    calculate the slope, m   as    (y1-y2) / (x1-x2)

       you will then have   y = mx + b

      then put one of the points into your building equation to calculate 'b'  

             (BTW you can 'see'  'b' where x = 0 the value of b is shown as -4 on the graph)

Sure. 

To find the range, we first need to set f(x) to y.
\[f(x) = \frac{x^2 + 14x + 9}{x^2 + 2x + 3}\]
\[y=\frac{x^2 + 14x + 9}{x^2 + 2x + 3}\]
Multiplying both sides by x^2 + 2x + 3 will get us:
\[\frac{x^2 + 14x + 9}{x^2 + 2x + 3}(x^2+2x+3)=y(x^2+2x+3)\]
\[\frac{x^2 + 14x + 9(x^2+2x+3)}{x^2 + 2x + 3}=y(x^2+2x+3)\]
\[x^2 + 14x + 9=y(x^2+2x+3)\]
Since the domain of this function is all real numbers, the discriminant has to be greater than or equal to 0. Solving for the discriminant of x^2 + 14x + 9=y(x^2+2x+3), we get:
\[x^2 + 14x + 9=y(x^2+2x+3)\]
\[x^2 + 14x + 9=yx^2+2yx+3y\]
\[x^2 + 14x + 9-(yx^2+2yx+3y)=0\]
\[(1-y)x^2 + (14-2y)x + 9-3y=0\]
\[(14-y)^2-4(1-y)(9-3y)\]
\[=196-56y+4y-(36-48y+12y^2)\]
\[=-8y^2-8y+160\]
Now that we have our discriminant, we can go on to solving our inequality, -8y^2-8y+160 ≥ 0.
\[-8y^2-8y+160 \geq 0\]
\[y^2+y-20 \leq 0\]
\[(y+5)(y-4) \leq 0\]
Using test intervals, we see that the inequality that works for y is -5 ≤ y ≤ 4
Therefore, our range is [-5, 4]

.....but the question states he cannot get anyone to do it !    So  no combos ?

we put them into groups like this 100-98=2 like hwat geno3141 said and we have 100/2=50 groups of 2 numbers of 2 so the sum is 2*50=100

Yeah, I agree.

I'm not really sure if it's 0 or 40 though

It is 40:

1 way for the first

2 for the second

5 for the third

4 for the last one

1*2*5*4=\(\boxed {40}\)

I thought the original answer was correct.....since no other one would do it and Tim had to do it himself.....there are no other combos ????/

6 + thirteen d's = 32

13d = 26

d = 2         

3rd term is 6   next is  8     10    12   14  16  18 20  22 24 26  28 30 32=16th term

1st term is   6 - 2 -2 = 2

Sum_n = n/2 (a₁ + a_n)

          = 20/2 (2 + 40) = 420

1t is 1*2*5*4=40 so there are 40 combinitions

40???

Alan is right of course but maybe you need to solve it algebraically?

It is also good to notice things such as the numerator  (looked at separately) has irrational roots

and the denominator is always positive. 

As x tends to plus or minus infinity f(x) tends to 1

I am not sure how to do it but I would not be cross multiplying like you have done.

The denominator is always positive, the minimum of the denominator is 2 at x=-1

When f(-1)=-2

For    \(-7-\sqrt{40}

The rest of the time f(x) is great or equal to 0.

I can also work out that (-0.5,1) is between the max and the min.

I have not answered your question.

I have just given you some initial skills to start graphing the curve.

Maybe I would try to do more with calculus not really sure and I'm unable to spend more time on it right now.

Aha! Yes, I have gotten the answer. Thanks to all who helped!

y  =  -2x² + bx + c          with roots     3 + sqrt(5)     and     3 - sqrt(5)

One answer is:  y  =  [ x - ( 3 + sqrt(5) ) ] · [ x - ( 3 - sqrt(5) ) ]

Let's multiply out the right-hand side:

     [ x - ( 3 + sqrt(5) ) ] · [ x - ( 3 - sqrt(5) ) ]     --->     [ x - 3 - sqrt(5) ] · [ x - 3 + sqrt(5) ]

                  --->   x² - 3x + sqrt(5)·x - 3x + 9 - 3·sqrt(5) - sqrt(5)·x + 3·sqrt(5) - 5

                  --->   x² - 6x + 4

However, since the original problem started with "-2x²", we'll have to multiply this answer by -2:

--->     y  =  -2x² + 12x - 8

To find the vertex:                              y  =  -2x² + 12x - 8

--->  add 8 to both sides:             y + 8  =  -2x² + 12x

--->   factor out the -2:                 y + 8  =  -2(x² - 6x)

--->   complete the square:    y + 8 - 18 =  -2(x² - 6x + 9)

                                                    y - 10  =  -2(x - 3)²  

The vertex is  (3, 10)

We look for the number of points in the triangle marked by the -axis, -axis, and the line . We consider the rectangle with the vertices (0,0), (5,0), (0,4), and (5,4) (cutting this rectangle diagonally results in our triangle). The points contained in the 4x5 rectangle are a 3x4 grid of points, or 12 points in total. Since none of the points lie on the diagonal of the rectangle and the triangle is half of the rectangle,  of the 12 points are contained in the triangle.