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let y = \(\sqrt[3]{x}\):

\(4y-\frac{2y^3}{y^2}=7+y\\4y-2y=7+y\\y=7\)

\(7=\sqrt[3]{x}\\x=343\)

NGL this is the dumbest question I've ever seen

You will need a few minutes time on a Cray supercomputer

1 + 1¹ =

1 + 1 = 2

36 points  / 3games   = 12 points per game

    (  or    3 games /36 points  = 1/12  game per point<======= this is not likely the one you you want though)

pi r^2 = original area

pi (2r)^2   =  pi r^2 + 108

4pi^2  pi   = pi r^2 + 108

3 pi r^2 = 108

r = 3.385                              new radius =6.77 units

Put into standard circle form :

(x +1)^2   +  (y-5.5)^2   = -8 +1+ 30.25              r = sqrt 23.25     center = -1, 5.5     

Minumum 'x'   will be     - 1 - sqrt 23.25

Minumum 'y' will be         5.5 - sqrt 23.5       (check my math)

As follows:

Your idea of using the inequality on AM and GM is correct.

Let $\alpha, \beta, \gamma$ be the real nonnegative roots of the given polynomial,
thus, $x^3+ax^2+bx+c = (x-\alpha)(x-\beta)(x-\gamma)$.  By Vieta's formula,
$\alpha\beta+\beta\gamma+\gamma\alpha = b$ and $\alpha\beta\gamma=c$.

Can you do the rest?

Note that $1/(abc) + 1/(abd) + 1/(acd) + 1/(bcd) = (a+b+c+d)/(abcd)$.  The values of $a+b+c+d$ and $abcd$ can be read off from the coefficients of $x^4-7x^3+4x^2+7x-4=0$ by Vieta's formula.
 

Completing the square on $x$ and $y$ independently to write the given expression as
$(x+a)^2+(y+b)^2+c$.   The answer is $c$.
 

52.3   -    1.2 degrees/thousand  * 2 thousand   =    52.3 - (1.2 * 2) = __________ degrees

https://web2.0calc.com/questions/quadratic-function_12

# = 10 000 * 2^(t/40)       Where  ' t '  is in minutes

b)  One hour = 60 minutes     plug that into the equation for 't'

c) # = 50 000+m       sub that into the equation and solve for 't'  (what is 'm' ?)

This question is triple posted :

https://web2.0calc.com/questions/quadratic-function_12

You question does not state what 'm' is....

    the function is a dome shaped parabola due to the  negative coeficient of   s^2

       the maximum (at the vertex) will be at a value of 's' given by     - b/ 2a   where    b = 3+m    and    a = - 1/28

            you will need to know the value of 'm' to calculate this speed

                use this speed in the equation to calculate the car's best mileage.... 

In triangle ABC, points D and E lie on BC and AC, respectively.

If AD and BE intersect at T so that AT/DT = 3 and BT/ET = 4, what is CD/BD?
 

\( \text{Let $\vec{CA}=\vec{b}$} \\ \text{Let $\vec{CB}=\vec{a}$} \\ \text{Let $\mu = \dfrac{3}{4} = \dfrac{AT}{AD}$} \\ \text{Let $1-\mu = \dfrac{1}{4} = \dfrac{DT}{AD}$} \\ \text{Let $\rho = \dfrac{4}{5} = \dfrac{BT}{BE}$} \\ \text{Let $1-\rho = \dfrac{1}{5} = \dfrac{ET}{BE}$} \\ \text{Let $\lambda = \dfrac{BD}{BC} $} \\ \text{Let $1-\lambda = \dfrac{CD}{BC} $} \\ \text{Let $\mathbf{ \dfrac{1-\lambda}{\lambda} = \dfrac{CD}{BD}} $} \\ \text{Let $\epsilon = \dfrac{CE}{CA} $} \)

\(\text{In C-D-T-E} \\ \begin{array}{|rcll|} \hline (1-\lambda)\vec{a}+(1-\mu)[\vec{b}-(1-\lambda)\vec{a}] &=& \epsilon \vec{b} + (1-\rho)(\vec{a}-\epsilon \vec{b}) \\ (1-\lambda)\vec{a}+(1-\mu)\vec{b} -(1-\mu)(1-\lambda)\vec{a} &=& \epsilon \vec{b} + (1-\rho)\vec{a} - (1-\rho)\epsilon \vec{b}) \\ \vec{a}[(1-\lambda)-(1-\mu)(1-\lambda)-(1-\rho) ] &=& \vec{b} [\epsilon - (1-\rho)\epsilon -(1-\mu) ] \\ \vec{a}[(1-\lambda)\Big(1-(1-\mu)\Big)-(1-\rho) ] &=& \vec{b} [\epsilon - (1-\rho)\epsilon -(1-\mu) ] \\ \vec{a}[\underbrace{ (1-\lambda)\mu)-(1-\rho) }_{=0} ] &=& \vec{b} [\underbrace{ \epsilon - (1-\rho)\epsilon -(1-\mu) }_{=0} ] \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline (1-\lambda)\mu-(1-\rho) &=& 0 \\ (1-\lambda)\mu &=& (1-\rho) \\\\ \mathbf{ 1-\lambda } &=& \mathbf{\dfrac{(1-\rho)}{\mu} } \\ \hline 1-\lambda &=& \dfrac{(1-\rho)}{\mu} \\ \lambda &=& 1-\dfrac{(1-\rho)}{\mu} \\ \mathbf{ \lambda} &=& \mathbf{ \dfrac{\mu-(1-\rho)}{\mu} } \\ \hline \dfrac{CD}{BD} &=& \dfrac{1-\lambda}{\lambda} \\ \dfrac{CD}{BD} &=& \dfrac{\dfrac{(1-\rho)}{\mu}} {\dfrac{\mu-(1-\rho)}{\mu}} \\ \\ \dfrac{CD}{BD} &=& \dfrac{(1-\rho)} {\mu-(1-\rho)} \\ \\ \dfrac{CD}{BD} &=& \dfrac{\dfrac{1}{5}} {\dfrac{3}{4}-\dfrac{1}{5}} \\ \\ \dfrac{CD}{BD} &=& \dfrac{\dfrac{1}{5}} {\dfrac{11}{20}} \\ \\ \dfrac{CD}{BD} &=& \dfrac{1}{5} * \dfrac{20}{11} \\ \\ \mathbf{ \dfrac{CD}{BD}} &=& \mathbf{\dfrac{4}{11}} \\ \hline \end{array}\)

For how many positive integers n less than 100 is a multiple of 6?

Lets simplify the expression (mod6)

\(5^n\mod6\\ \equiv(-1)^n\mod6\\ \)

which is -1 when n is odd and +1 when n is even

\(13^{(n+2)}\equiv 1^{(n+2)}\equiv1\mod6\)

\(8^{(n+1)}\equiv 2^{(n+1)}\equiv2*2^n\mod6\\~\\ 14^{(n+3)}\equiv 2^{(n+3)}\equiv8*2^n\equiv2*2^n\mod6\\~\\ 8^{(n+1)}+14^{(n+3)}=4*2^n\mod6\\~\\ \)

so

\(5^n + 8^{(n + 1)} + 13^{(n + 2)} + 14^{(n + 3)}\mod6 \\ \equiv 1+ (-1)^n+4*2^n \mod6\\~\\ \text{When n is odd}\equiv 4*2^n \mod6\\~\\ \text{When n is even}\equiv 2+4*2^n \mod6\\~\\ \)

consider 4*2^n

So that expression is a multiple of 6 for all even values of n.  So that is 49 values,  2 to 98 inclusive

y=0 basically means the x axis.

if it wasnt for the\( f(x)+g(x)\) condition, you could just chuck a negative sign in front of it and you were good to go.

anyway -- we are given that   \( f(x) = \frac{(3x^4+3x^3+3x^2+3)} {(x^2+2x-4) } \) -- now what is the value of the polynomial g(x) so that it satisfies the condition? lets set up this 'equation' : 

 \( \frac{\left(3x^4+3x^3+3x^2+3\right)}{\left(x^2+2x-4\right)}+g(x)=\frac{-\left(3x^4+3x^3+3x^2+3\right)}{\left(x^2+2x-4\right)}\)  

 \(g(x)=-\frac{3x^4+3x^3+3x^2+3}{x^2+2x-4}-\frac{3x^4+3x^3+3x^2+3}{x^2+2x-4} \) 

 \(g(x)=-\frac{2\left(3x^4+3x^3+3x^2+3\right)}{x^2+2x-4} \) 

lets confirm the condition:

 \( f(x)+g(x)\) \(\Rightarrow\) \(\frac{(3x^4+3x^3+3x^2+3)} {(x^2+2x-4) } +-\frac{2\left(3x^4+3x^3+3x^2+3\right)}{x^2+2x-4}  = \boxed{\frac{-3x^4-3x^3-3x^2-3}{x^2+2x-4}}\) 

if you graph it you would get:

: 

here is the link for it as well: https://www.desmos.com/calculator/v2w0ukjkoq

:D

CD / BD = 4/11  

5*25*26

This graph should help

As follows:

If you make a table of congruences modulo 6 of

$$5^n, 8^{n+1}, 13^{n+2}, 14^{n+3)$$

for $n \equiv 0,1,2,3,4,5\pmod6$, you'll see a pattern.  According to my table, half of $$5^n + 8^{n+1} + 13^{n+2} + 14^{n+3)$$ for $1\le n< 100$ are $\equiv 0\pmod6$ and the other half are $\equiv 2\pmod6$.

For conveniece, when creating the table, replace every base $a$ by $a \bmod 6$.
 

when you're trying to find the greatest product, you want to have two numbers that are the closest possible distance to each other.

in this case, the pair would be 12 and 12, so the greatest product is simply 144.

to verify this answer, you can try out different pairs.

for example, let's try 13 and 11. 

13 * 11 = 143, which is less than 144. you can continue trying other pairs if you are still doubtful ;)

hope this helped! please let me know if you are confused about anything i did