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hmm...that's weird. do you get that when you enter an equation in the calculator and click the equal sign to solve?

-1 x 10 = -10

-1 + 10 = 9

I hope this helps 

I keep getting this 

Figured it out myself, this is an interval problem not a quadratic problem, btw you were wrog the answer was -7.

the calculator for web2.0? it works fine for me

By Vieta's formulas, b + c = (-2) + 3 + (-2)(3) = -5.

nvm i got it

y = 31 - 5x. Sub that into second equation we get 3x + 124 - 20x = -37. -17x = -161, so x = 161/17. Sub that into y = 31 - 5x we get y = 31-5*161/17 = 31 - 805/17 = -278/17. 

(161/17, -278/17)

Find the unique pair of real numbers (x,y) satisfying 

(3x^2 + 6x + 5)(2y^2 + 8y + 10) = 4.

(3x^2 + 6x + 5)*2(y^2 + 4y + 5) = 4.

(3x^2 + 6x + 5)(y^2 + 4y + 5) = 2

try

3x^2 + 6x + 5= 2           and       y^2 + 4y + 5=1

3x^2 + 6x + 3= 0           and       y^2 + 4y + 4= 0

3(x^2+2x+1)=0             and         (y+2)^2=0

3(x+1)^2=0                   and         (y+2)^2=0

x=-1,   y=-2

This is one solution.  I do not know how to show algebraically that it is unique though.  

Here is an interactive graph you can play with if you want.  Graphically, the solution is shown to be unique.

$either\\\quad x^2-5x+2 = 0 \quad and \quad x^2-4x+2\ne0\\ \quad x=\frac{5\pm\sqrt{17}}{2} \quad and \quad x\ne\frac{4\pm \sqrt{8}}{2}\\ \quad x=\frac{5\pm\sqrt{17}}{2}\\ or\\ \quad x^2-4x+2=1\\ \quad x^2-4x+1=0\\ \quad x=\frac{4\pm\sqrt{12}}{2}=2\pm\sqrt{3}\\ sum=\frac{5+\sqrt{17}}{2}+\frac{5-\sqrt{17}}{2}+2+\sqrt{3}+2-\sqrt{3}\\ sum=\frac{5}{2}+\frac{5}{2}+2+2\\ sum=9$

Note. I have not checked this answer, there could easily be careless errors.

LaTex:

either\\\quad x^2-5x+2 = 0   \quad and \quad x^2-4x+2\ne0\\
\quad x=\frac{5\pm\sqrt{17}}{2} \quad and \quad x\ne\frac{4\pm \sqrt{8}}{2}\\
\quad x=\frac{5\pm\sqrt{17}}{2}\\
or\\
\quad x^2-4x+2=1\\
\quad x^2-4x+1=0\\
\quad x=\frac{4\pm\sqrt{12}}{2}=2\pm\sqrt{3}\\
sum=\frac{5+\sqrt{17}}{2}+\frac{5-\sqrt{17}}{2}+2+\sqrt{3}+2-\sqrt{3}\\
sum=\frac{5}{2}+\frac{5}{2}+2+2\\
sum=9

Suppose we have $PQ=6$, $QR=7$, and $PR=9$, as in the picture below:

Find the length of the median from $R$ to $\overline{PQ}$.

cos Rule:

$\begin{array}{|rcll|} \hline 9^2&=&(2*3)^2+7^2-2*2*3*7*\cos(Q) \\ 9^2&=&6^2+7^2-2*2*3*7*\cos(Q) \\ 2*2*3*7*\cos(Q) &=& 6^2+7^2-9^2 \\ 2*2*3*7*\cos(Q) &=& 4 \quad | \quad :2 \\ 2*3*7*\cos(Q) &=& 2 \qquad (1) \\ \hline \end{array}$

cos Rule:

$\begin{array}{|rcll|} \hline d^2&=&3^2+7^2-2*3*7*\cos(Q) \\ 2*3*7*\cos(Q) &=& 3^2+7^2-d^2 \\ 2*3*7*\cos(Q) &=& 58-d^2 \qquad (2) \\ \hline \end{array}\\ \begin{array}{|lrcll|} \hline (1)=(2):& 2 &=& 58-d^2 \\ & d^2&=& 58-2 \\ & d^2&=& 56 \\ & d^2&=& 4*14 \\ & \mathbf{d}&=&\mathbf{ 2\sqrt{14} } \\ \hline \end{array}$

Depending on where you mean the cot(5x) to go, one of these graphs should help.

Sorry, There is a typo!

$\mathbf{PM} = \mathbf{ 2\sqrt{37} }$

Alright. So we know that the interior angles of a triangle add up to 180 degrees, so we find the interior angles of each angle by subtracting the exterior angle from 180.

$1: 180-158=22$

$2:180-99=81$

Therefore, 22+81+c(the third angle) must be 180.

103+c=180

$\fbox{c=77}$

PR = a = 9          QR = b = 7           PQ = c = 6          RM = d = ?

d = 1/2[sqrt(2*a² + 2*b² - c²)]

d = √224 / 2 ≈ 7.483

For future reference, the measure of the angles in an equaingular polygon with n sides is $(n-2)\cdot180\over n$.

3x + 62 = 5x + 60

2x = 2

x = 1

AB = 65

AC = 65

BC = 37

Perimeter = 167

You've got the wrong answer!!!

-------------------------------------------

AN is the height of triangle ABC.

Angle ANC = 90 degrees

Angle C = 60 degrees

Therefore, AN = sin(60) = 0.866

btw, CN = 0.433

It's actually 30 feet shorter. Don't worry. Just a mere typo.  

1 * (5 + 8) = 13

1 + (5 * 8) = 41

If a, b, and c are positive integers satisfying ab + c = bc + a = 13  and ac + b = 41, what is the value of a + b + c?

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

b = 1

ac + 1 = 41

ac = 40

a + c = 13

(40/c) + c = 13

c = 8,   c = 5

a + b + c = 14

In for the gruelling run for step by step....

for  $1 \star 2$, we can list it as

$\frac{1+2}{1-2}=\frac{3}{-1}=-3$

and for $-3 \star 3$ we can write it as:

$\frac{0}{-6}$

or, just $\fbox{0}$