NotThatSmart

First, we can use variables to solve this problem in  quite a handy way. First, let's set

\(x = 2.\overline{57}\)

For this value of x, we know that

\(100 x= 257.\overline{57}\)

Now, we do the really cool trick. We subtract the first equation from the second equation to get

\(100 x - x = 257.\overline{57} - 0.\overline{57}\)

Notice that the repeating terms cancel out. Thus, we have

\(99x =257\\ x= \frac{255}{99}\)

Dividing by a common factor, we have

\(x= 2 \frac{19}{33}\)

Thus, our final answer is \(2.\overline{57} = 2 \frac{19}{33}\)

Thanks! :)

First, let's combine all like terms and see what we get. We find that

\(y=-5x^{2}-6x+10\)

Now, the x value of the vertex can be written in the equation \(-b/2a\) for quadratics in the form \(ax^2+bx+c\)

Thus, we have

\(x=-(-6)/2(-5) = -3/5\)

Plugging this in to find the y value, we get

\(y=11.8\)

Thus, the vertex is at \((-0.6, 11.8)\)

So our answer is \((-0.6, 11.8)\)

Thanks! :)

Let's first combine all like terms and see what we get. We have

\(x^{2}+18x+3\)

Now, we simply complete the square for x. Adding and subtracting 81, we get

\(x^2+18+81-81+3\\ (x+9)^2-78\)

Thus, the first blank is 1, the second blank is 9, and the third blank is -78. 

So that is our final answer. 

Thanks! :)

Let's first combine all like terms and simplify the equation. We have

\(x^{2}+15x+k\)

Now, there are many ways for this equation to be a double root. 

As long as two factors of k add up to 15, we are good. 

For example, we have

\(k=36; x^2+15x+36 = (x+12)(x+3)\\ k=14; x^2+15x+14 = (x+14)(x+1)\\\)

Negative k values also work. For example. we have

\(k=-54; x^2+15x-54= (x+18)(x-3)\\\)

Thanks! :)

Let's set the total value to be r. We have that

\(r = \sqrt{x - \sqrt{x}}\)

Squaring both sides, we get 

\({r}^{2} = x - \sqrt{x}\)

Now, let's note something important. Since both must be integers, we find that x must be a perfect square. 

Let's set \(x={a}^{2}, a\ge0\)

Subsituting this in, we find that

\({r}^{2}={a}^{2}-a\)

Let's notice that this isn't possible. There isn't a three digit value of x that satisfies this equation. 

Thus, our answer is none. 

I can elaborate if needed. 

Thanks! :)

First, let's note that triangle ABC is a 45-45-90 triangle. 

Thus, we can write

\(AB = BC = 12/\sqrt 2 = 6\sqrt 2 \)

Also, let's note that from the problem, we have thart

\(AD = AD\\ BD = BD\\ AB = BC\)

This means that triangles ABD and CBD are congruent. 

Thus, we have

\([ ABD ] = (1/2) [ABC] = (1/2) ( 1/2) ( 6\sqrt 2)^2 = (1/4) (72) = 18\)

Thus, our answer is 18. 

Thanks! :)

We can write an inequality to solve this problem. We have

\(x^2 + (17 - m)x + 4 > 0\)

The rule states that if there are two distinct roots, then the descriminant must be greater thnan  0. Thus, we have the equation

\((17 - m)^2 - 4(1)(4) > 0 \\ (17 - m)^2 > 16\)

We take both roots from the equation and solve both. We have

\(17 - m > 4 \\ 17 - 4 > m \\ m < 13 \)

solving the other root, we get

\( 17 - m < -4 \\ 21 < m \\ m > 21 \)

Thus, in interval notation, we have

\(( -\infty , 13)U ( 21, \infty)\)

So \(( -\infty , 13)U ( 21, \infty)\) is our final answer. 

Thanks! :)

We have a system of equations we need to solve. 

We have the system

\(x=4\\ x^2+y^2=25\)

Plugging in x as 4 into the second equation, we have

\(16+y^2=25\\ y^2=9\\ y= \pm 3\)

Thus, we find there are two ordered pairs and solutions to the equation. We have that

\((4, 3); (4, -3)\)

The distance between the two is just \(3-(-3) = 6\) since they share the same x value. 

Thus, 6 is our answer. 

Thanks! :)

First, let's combine all like terms for x and move all terms to one side. We have that

\(x^{2}+12x-4=0\)

Using the quadratic equation, we have that

\(x=\frac{-12\pm \sqrt{12^{2}-4\cdot 1(-4)}}{2\cdot 1}\)

\(x=2\sqrt{10}-6\\ x=-2\sqrt{10}-6\)

Now, we can use a handy trick to solve this problem. We have that

\((2\sqrt{10}-6)^2+(-2\sqrt{10}-6)^2\)

Expanding and distributing in our values, we have

\(-24\sqrt{10}+76+4\left(\sqrt{10}\right)^{2}+12\sqrt{10}-6\left(-2\sqrt{10}-6\right)\)

All the radicals basically cancel out. In the end, we are left with 152. 

So 152 is our answer. 

Thanks! :)