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f(2)  =  4² - 1   =   16 - 1   =   15           At  2  hours, there are  15  students enrolled.

f(4)  =  4⁴ - 1   =   256 - 1   =   255       At  4  hours, there are  255  students enrolled.

The number of students goes from  15  to  255 .

255 - 15  =  240      So the number of students increased by  240 .

The number of hours goes from  2  to  4 .

4 - 2  =  2      So the number of hours increased by  2 .

In  2  hours time, the number of students increased by  240 .

On average, that is an increase of  240/2  ,  or  120 ,  students per hour.

The average increase in the number of students per hour is  120  .

We know

d   =   r * t              Divide both sides of this equation by  r  to get

d / r   =   t

So....

t  =  d / r

Is something missing from the question?

11/25  does not equal  4/5  .    4/5  is bigger.

So...   11/25  =  4/5   this isn't true

Glad you got them figured out!! Good luck getting that  "A" !!    💯

\(\frac{59}{7}-[(x+\frac{30}{7})-\frac{22}{7}]=\frac{19}{7} \\~\\ \frac{59}{7}-[x+\frac{30}{7}-\frac{22}{7}]=\frac{19}{7} \\~\\ \frac{59}{7}-x-\frac{30}{7}+\frac{22}{7}=\frac{19}{7} \\~\\ \frac{59}{7}-\frac{30}{7}+\frac{22}{7}=\frac{19}{7}+x \\~\\ \frac{59}{7}-\frac{30}{7}+\frac{22}{7}-\frac{19}{7}=x \\~\\ \frac{32}{7}=x\)

the slope between two points   \(=\,\frac{\text{difference in y values}}{\text{difference in x values}}\)

the slope between  (2, 4)  and  (6, 3)   \(=\,\frac{3-4}{6-2}\)

the slope between  (2, 4)  and  (6, 3)   \(=\,-\frac{1}{4}\)

And we know that the slope between  (6, 3)  and  (-5, c)  also equals  -\(\frac14\) .  So we know

\(-\frac14\,=\,\frac{c-3}{-5-6}\\~\\ -\frac14\,=\,\frac{c-3}{-11}\)

                                              Multiply both sides of the equation by  -11 .

\(\frac{11}4\,=\,c-3\)

                                              Add  3  to both sides.

\(\frac{11}4+3\,=\,c \\~\\ c\,=\,\frac{23}{4}\,=\,5.75\)

Let's call the larger number  L , and the smaller number  S .

1.     L  =  S + 4\(\frac23\)

2.     L - S  =  S

                                               Substitute  S + 4\(\frac23\)  in for  L .

S + 4\(\frac23\)  -  S  =  S

                                               Cancel the  S  and  -S  on the left side.

\(4\frac23\)  =  S

L  =  S + 4\(\frac23\)

                                               Substitute  4\(\frac23\)  in for  S .

L  =  4\(\frac23\)  +  4\(\frac23\)

L  =  8 +  \(\frac43\)

L  =  8 +  \(\frac33\) +  \(\frac13\)

L  =  9\(\frac13\)

And...let's check that their sum is a natural number.

L + S   =   9\(\frac13\)  +  4\(\frac23\)   =   13 \(\frac33\)   =   14   Yep!  

(x + 3)²  +  (y - 3)²  =  6          Let's solve this for  y .

(y - 3)²   =   6 - (x + 3)²

 y - 3   =   ±√[ 6 - (x + 3)² ]

 y   =   ±√[ 6 - (x + 3)² ]  +  3           So....using this value for  y....

\(\frac{y}{x}\,=\,\frac{\pm\sqrt{6-(x+3)^2}+3}{x}\)

We can say

\(Y\,=\,\frac{\pm\sqrt{6-(x+3)^2}+3}{x}\)      and we want to know the maximum Y value here.

First let's find the slope between these two points.

slope  \(=\,\frac{\text{change in y}}{\text{change in x}}\,=\,\frac{5-(-16)}{4-(-3)}\,=\,\frac{5+16}{4+3}\,=\,\frac{21}{7}\)   =   3

Using a slope of  3  and the point  (4, 5) , the equation of the line in point-slope form is

y - 5  =  3(x - 4)          Distribute the  3 .

y - 5  =  3x - 12          Add  5  to both sides.

y  =  3x - 7