NotThatSmart

First, let's simplify all like terms and bring all terms to one side. 

We get the equation

\(2x^{2}+2x+7=0\)

Now, using the quadratic equation, we finally get two answers. We find that we have

\(x=-\frac{1}{2}+\sqrt{13}\cdot (\frac{1}{2}i)\\ x=-\frac{1}{2}-\sqrt{13}\cdot (\frac{1}{2}i)\)

Now we know what A and B are. We find that

\(a=-1/2; b = \frac{\sqrt{13}}{2}\)

Multiplying them toeghet, we finally come up with the answer. 

We finally have 

\(ab=-\frac{\sqrt{13}}{4}\)

Thus, our final answer is -sqrt(13)/4

Thanks! :)

First, let's focus on the expression we need A and B to be equal to. 

In order to find what we need, we need to be able to compare coefficients. let's convert it to standard form. 

We have

\((x - 2)(x + 2)=x^2-4\)

However, the constans don't match. Thus, we need to divide what we have by -4 to get a constant of +1. 

We have

\(-\frac{1}{4}x^2+1\)

Now. we cam compare coefficients. Since A is the coefficient of x^2, we have

\(A=\frac{-1}{4}\)

Since B is the coefficient of x, we have \(B=0\)

Thus, we finally have \(A+B=-\frac{1}{4}+0 = -\frac{1}{4}\)

So -1/4 is our final answer. 

Thanks! :)

First, let's establish the equation between Farenheit and Kelvin. 

We have 

\(F = \frac{9} {5} ( K − 273 ) + 32\)

Now, since we want them to be equal, let's set a varianle. 

Let's let K and F both be X. Thus, we have the equation 

\(X = \frac{9} {5} ( X − 273 ) + 32\)

So, after we solve for X, we get that

\(X=574.25\)

So 574.25 is our answer. 

Thanks! :)

We can set variables to solve this problem. 

First, let's let the reciprocals be \( 1/x ,1/y \)

The arithmetic mean of the two numbers are \( [ 1/x + 1/y] / 2 = [ x + y] / [ 2xy]\)

This means the harmonic mean is \( [ 2xy] / [ x + y] \)

Now, we write an equation. Since the harmonic mean must eqaual 20, we set the equation up. 

Now, we put x in terms of y. We have

\(2xy / [ x + y ] = 20 \\ xy / [ x + y] = 10 \\ xy = 10x + 10y \\ yx - 10y = 10x y ( x - 10) = 10x \\ y = 10x / [ x - 10]\)

I counted 7 numbers. Not sure if that's right. 

Thanks! :)

First, let's simplify and combine all like terms. 

After doing that, we get

\(3x^2-kx+33=0\)

Now, in order for the equation to have no real solutions, the descriminant must be less than 0. 

The descriminant is \(b^2-4ac\), so we have the equation

\(k^2 - 4(33)(3) < 0\)

Since we are trying to find the largest, we set the two to equal each other and get

\(k^2 - 396 = 0\\ k^2 = 396\\ k \approx 19.89974\)

Rounding down, the largest k can be as an integer is 19. 

So our answer is 19. 

Thanks! :)

First, let's combine all like terms. 

By setting up a quadratic, we get \(3x^{2}+4x+10\)

The descriminant of a quadratic can be written in the form \(b^2-4ac\) for quadratcis written in the form \(ax^2+bx+c\)

Thus, we have

\(4^2-4(3)(10)\\ 16-120\\ -104\)

So -104 is our answer

Thanks! :)

The slope of line PQ is

\(m_{PQ}=\frac{y_q-y_p}{x_q-x_p}=\frac{4.9029-4.8507}{0.9806-1.2127}\)

\(m_{PQ}=-0.2249\)

-0.2249 is the answer.