Two parabolas are the graphs of the equations $y=2x^2-10x-10$ and $y=x^2-4x+6$. give all points where they intersect. list the points in order of increasing $x$-coordinate, separated by semicolons.

(-2, 18) and (8, 38)

Step-by-step explanation:

First, set the two equations equal to each other to get $2x^2-10x-10=x^2-4x+6$. Combine like terms to get $x^2-6x=16$. To complete the square, we need to add $\left(\dfrac{6}{2}\right)^2=9$ to both sides, giving $(x-3)^2=16+9=25$.

So we have $x-3=\pm5$. Solving for $x$ gives us $x=-2$ or $8$. Using these in our original parabolas, we find the points of intersection to be $\boxed{(-2,18)}$ and $\boxed{(8,38)}$.

Credit: AoPs

1.true 2.false? ? i think. .

step-by-step explanation:

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116