NotThatSmart

We need to note a couple of important points to find the equation of this parabola. 

We need one random point along the parabola and the vertex. 

Let's choose the point (0, 7) since it's the y intercept, and the vertex of the graph is \((3, 1)\)

Now, since we have the vertex of the graph, we can put this equation into intercept form. 

We have

\( f(x) = m ( x-3)^2 + 1\) where m is a real number. 

Now, we simply subsitute in the point we retrireved, (0, 7), and plug it in the find m. 

We have that

\(7 = m(0-3)^2+1\\ 7 = 9m+1\\ m = 2/3\)

Now that we have m, we have the full equation of the parabola in INTERCEPT form. 

We need to convert it into standard form to find a,b,c. 

The equation is now

\( f(x) = 2/3(x-3)^2 + 1 \)

This expands to the equation

\(2/3 x^2 -4 + 7 \) meaning that we found each value of a,b, and c. 

We have that \(a = 2/3 , b = - 4 , c = 7 \). Multiplying these 3 together, we get \(2/3*(-4)*7 = -28 *2/3 = -56/3\)

So -56/3 is our final answer. 

Thanks! :)

I can only think of two combinations I can think of are

Combinations of \(2, 2, 4, 5, 5, 5\) and combinations of \(1, 4, 4, 5, 5, 5    \)

Now, we can calculate the number of ways or organize each combination and add them together. 

The number of permutations of 2, 2, 4, 5, 5, 5 is \( 6! = 720 \)  

The number of permutations of 1, 4, 4, 5, 5, 5 is \(6! = 720\)

Now, we had these together to get

\(720 + 720 = 1440 .\)

So I think this is the correct answer, but I'm not actually very sure. 

Thanks! :)   

First, we can write a handy equation to solve for the constant and find the 3 terms. 

Let's let the constant added to every number be x. 

Since the three terms for a geometric series, we have the equation

\(\frac{100+x}{60+x} = \frac{140+x}{100+x}\)

Now, when we crossmultiply and then expand everything out, we get

\((x+100)(100+x)=(x+60)(140+x)\\ 200x+x^{2}+10000=140x+x^{2}+{8400+60x}\\ 200x+x^{2}+10000=200x+x^{2}+8400\)

Now, we bring all terms to one side of the equation. We have

\(200x+x^{2}+10000-x^{2}-200x-8400=0\\200x+1600-200x=0 \\ 1600=0\)

However, this statement is obviously not true, meaning that x is invalid. 

This also means that there are NO solutions to this given problem. 

*Note, I may have made a mistake. Not sure. 

Thanks! :)

First of all, let's focus on the second equation. 

We can factor the right side of the equation to get \((x+y)(x^2+xy+y^2)=162+x^2+y^2\)

Since x+y is equal to 10 from the first equation, the equation becomes

\(10(x^2+xy+y^2)=162+x^2+y^2\)

\(10x^2+10xy+10y^2= 162+x^2+y^2\)

\(9x^2+9y^2+10xy=162\)

Now, isolating x in the first equation, we get

\(x=10-y\)

Now plugging this value of x into the second equation, we get

\(9(10-y)^2+9y^2+10(10-y)y=162\\ 9(100-20y+y^2)+9y^2+100y-10y^2=162\\900-180y+9y^2+9y^2+100y-10y^2=162\\ 8y^2-80y+900=162\\8y^2-80y+738=0\)

However, plugging this into the quadratic formula, notice that unofruntately, the descriminant is less than 0, menaing the eventual solutions are NONREAL numbers. 

Thanks! :)

If angle XWZ = 60°, angle WXY = 180° - 60° = 120° because XY and WZ are parallels.

However, this means that Angle ZXW must be less than 120°.

In the problem though, it clearly defines ZXW as 120, meaning this problem is invalid. 

Thanks! :)

We could probably use casework to solve this problem.

The first case to look for is only two digit numbers. 

We have 23 and 32, so there are 2 possibilities. 

Next, consider 3 digit numbers. 

We have 311, 131, 113 for 3 more. 

We also have 221, 212, and 122 to add 3 more on. 

Next, consider 4 digit numbers. 

The only numbers that work are numbers with three 1s and one 2. There are 4 ways we could do this. 

Lastly, we have 5 digit numbers. The only one we have to consider is \(111111\), one possibility. 

Now, we add up the numbers. We get \(2+3+3+4+1 = 13.\)

So 13 is our answer!

Thanks! :)