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Thank you 

Yes, Good work!   

[-6,-4)  and  (0,6) 

\(\sqrt{400} = 20\)

Every number from 1 - 400, so \(\color{brown}\boxed{400}\) integers

You could apply AM-GM to get

\(\frac{(y-x)^2}{(y-z)(z-x)} + \frac{(z-y)^2}{(z-x)(x-y)} + \frac{(x-z)^2}{(x-y)(y-z)}\ge3 \sqrt[3]{\frac{(y-x)^2}{(y-z)(z-x)} \cdot\frac{(z-y)^2}{(z-x)(x-y)} \cdot\frac{(x-z)^2}{(x-y)(y-z)}}\\=\boxed{3}\)

but idk if that's correct because it says real numbers not positive reals

\(1 - ({1 \over3}+{1\over4}+{1 \over4})\)

This equation leads you to the answer. If you still can't find it, use a calculator and evaluate the expression. 

To get the percentage, multiply your answer by 100. 

For the third question, they should give up swimming. 

The perimeter of the shaded region is  \(10+\text{arc}(QP)\). The circumference of the entire circle is \(10\pi\). The arc is \(1 \over 6\) of the total perimeter, meaning that the arc's length is \({5 \over 3}\pi\). Thus, the perimeter is \(\color{brown}\boxed{10+{5 \over3}\pi}\)

Then it should be (0,6)?

The question is actully asking .  "For what values of of x is the y values of the graph negative."

Well, it is negative between  x= -6 and x= 4      [-6,-4)     

The brackets are curly at x=-4 because at that point y=0 which is not zero.

NOW can you tell me what the other correct intervals are?

You got the right answer but what you have written is not altogether correct.

You have to treat the floor funtion like you would brackets.

So you work out what is inside and THEN take the flor of it befrome moving on to the rest of the expression.

I also desctribed what the floor funtion does rather badly.

The floor funtion simply ROUNDS A NUMBER DOWN to the nearest INTEGER.   

(Thanks for the guidence EP) 

I know you already worked out much of this but I wanted to reiterate the whole lot.  ;)

\(f(1)=\lfloor 1 \rfloor -2 = 1-2=-1\\ f(1.5)=\lfloor 1.5 \rfloor -2 = 1-2 = -1\\ f(3)=\lfloor 3 \rfloor -2 =3-2=1\)

giving    (1, -1)  (1.5, -1) and  (3, 1)

(1/2) lb per pizza  *  $9.91 / lb  * 36 pizzas  =

(1/2) (36) (9.91)   = 

18  * 9.91   =

$178.38

finally i can breath....

i've been holding my breath for 10 minutes now

THANK YOU CPHILL for adding proof to my solution. Also thanks to @Builderboi for providing an answer. Thanks ya'll!

Call T  the tuition now

Under the Dean's plan the  amount  after two years can be expressed as    T (1.10)^2 = 1.21T

The Provost's plan is a  little trickier .....   T +  ( T-10000).30   = 1.3T -3000 expresses the tuition after the first year

And   [ 1.3T - 3000] + [1.3T -3000 - 10000].30  =  1.69T -6900  expresses the tution after the second

Set  these equal

1.69T - 6900  = 1.21T

1.69 T - 1.21T = 6900

.48T = 6900

T= 14,375 =  current tuition  ( just  as XxmathguyxX   found !!! )

Proof

                                    Dean               Provost

After Year 1               15812. 50        15687.50

After Year 2               17393.75         17393.75         

GREAT JOB, XxmathguyxX    !!!!!

P.S  - you can  take a breath now.....

let's see what cphill gets.....

*getting so nervous*

*on the brink of extiction*

*super fast breathing*

First, I simplify \(\frac{ {\left( \frac{1}{2} \right)}^2 + {\left( \frac{1}{3} \right)}^2 }{ {\left( \frac{1}{4} \right)}^2 + {\left( \frac{1}{5} \right)}^2} = \frac{13x}{41y} \) so that it looks like \(\frac{ {\left(\frac{1}{25} \right)} + {\left( \frac{1}{9} \right)} }{ {\left( \frac{1}{16} \right)} + {\left( \frac{1}{25} \right)}} = \frac{13x}{41y} \), and then I added the fractions on the numerator and the fractions on the denominator so I got \((\frac{13}{36})/(\frac{41}{400})\), which can then be simplified to \(13*100/41*9\).

remember that this is equal to \(13x/41y\), so next, I get rid of 13/41 on both sides, giving me \(x/y=100/9\). I want \(\sqrt{x}/\sqrt{y}\), so I square root both sides of \(x/y=100/9\), which gets me\(\sqrt{x}/\sqrt{y}=10/3\).

I'm not really good at expressing things with the LaTeX thing, so I apologize if my answer is unclear, I can try it if you want.

i see that cphill is writing an answer currently

The first thing that I did was simplify ((1/2)^2+(1/3)^2)/((1/4)^2+(1/5)^2), which is (13/36)/(41/400). This can be simplified so that it looks like this: (13/41)(100/9). x/y=100/9, so \(\sqrt{x}/\sqrt{y}=10/3\)

also 144444 is indeed a very expensive school

Using implicit differentiation  we can find the slope at any point on the  ellipse

2x  + (1/4)(2))yy'  =  0

y' =     -2x / [ (1/2) y ]    =   -4x / y

We know the slope of the line   =3

So

-4x / y  =  3

-4x = 3y

(-4/3)x = y           

 (16/9)x^2 = y^2       sub this  back into the  equation of the ellipse

x^2  + (1/4)(16/9)x^2  =  1

x^2 + (4/9)x^2  = 1

(13/9)x^2  = 1

x^2  =  (9/13)              take both roots

x=  +3/sqrt 13     and x  = -3/sqrt 13

When x = 3/sqrt 13 , y=  (-4/3)x  = (-4/3) (3/sqrt 13) =  -4/sqrt 13

When x = -3/5 , y =  (-4/3)x = -(4/3)(-3/sqrt 13)  = 4/sqrt 13

So  the line is tangent to the ellipse at   ( 3/sqrt 13, -4/sqrt 13)   and  (-3/sqrt 13, 4/sqrt 13)

Since b is the y interecept in both cases let  (0, b)  be the b we are looking for

And we can use the slope formula to find b  in  each case

( b - -4/sqrt 13) / ( 0 - 3/sqrt 13)  = 3                            

(b + 4/sqrt 13) / ( -3/sqrt 13) = 3

b + 4/sqrt 13  = -9sqrt 13

b  = -13/sqrt 13 =  -sqrt 13

And by symmetry,  b = sqrt 13  is the other value for b

Here's a graph : https://www.desmos.com/calculator/kaztwjaevr

Percentages

Let's call X the original tuition, and let's assume it's greater than 10,000 so far. We'll analyze the first option:

Dean of Admissions suggests increasing tuition by 10% each year for the next two years. The first year the tuition would be. \(1.1x\)

The second-year the tuition would increase another 10% of the last amount, so it would end being. \(1.21x\)

Now for the second option:

The Provost suggests that each year they increase the tuition by 30% of the amount of the tuition above $10,000. The first increase would be \(0.3x-3000\)

And the new tuition at the end of the first year would be \(1.3x-3000\)

This new amount would be increased by applying the same rule, so the second-year increase would be 0.39x-3900

And the final tuition would be

1.69x - 6900

The Treasurer points out that the tuition will be the same under any scheme, so 1.69x - 6900= 1.21x

solve:

The actual tuition is $14,375

Since the focus lies above the directrix, this parabola opens upward

The vertex   will  be   (-2,  (-6 + -8)/2 )  =  (-2, -14/2 )  = ( -2,-7)

We have  the form  y =  [ 1/ (4p) ] ( x - h)^2  + k

(h,k)  = the vertex = (-2, -7)

p =  the dstance  between the  vertex and focus  ( or between  the  vertex and the diectrix)  =1

So....filling in what we know

y = [ 1/ (4*1) ] ( x - - 2)^2  - 7

y = (1/4) ( x + 2)^2 -7

You are correct  !!!

The answer to the second one is  0, 2

Write is as an equation, where \(x\) is the cost per year: 

\(1.21x=1.3(x-10,000)\)

\(1.21x=1.3x-13,000\)

\(-0.09x=-13,000\)

\(0.09x=13,000\)

\(\color{brown}\boxed{144,444 {5 \over 9}}\)

A very expensive state college indeed :)