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For

x+y+z=17, 

xy+xz+yz=94,

xyz=168

you can look at all the factors of 168

We want two quadratics with the same constant and degree. So let's try converting the quadratic we are given. 

First, let's combine all like terms to the quadratic we are comparing to. We find that

\(2x^2 - 5x + 6 \)

Let's note the constant is different. Thus, we must divide this by 6 so that we have the same constant as Ax^2+Bx + 1. We have

\((1/3)x^2 - (5/6) x + 1\)

Thus, we know what A and B are. We simply plug those in and get

\(A + B = (1/3) - (5/6) = (2/6) - (5/6) = -3/6 = -1/2 \)

So -1/2 is our final answer. 

Thanks! :)

Alright. First, let's combine all like terms to this horrible mess. We get that

\(6x^{2}-21x-31=0\)

Now, we simply apply the quadratic formula and take the smaller root. Using the fomrula, we get

\(x=\frac{21\pm \sqrt{(-21)^{2}-4\cdot 6(-31)}}{2\cdot 6}\)

From this monstrosity, we get that

\(x=\frac{\sqrt{1185}+21}{12}\\ x=\frac{-\sqrt{1185}+21}{12}\)

Obviously, the first term of x is bigger, so we take that as our answer. 

We find that x is \(x=\frac{\sqrt{1185}+21}{12}\)

So our answer is just the biiger root. 

Thanks! :)

also, 500th answer...woop woop! :)

First, let's note something important, 

The descriminant of the quadratic is \(b^2-4ac\) for any quadratic in the form of \(ax^2+bx+c\)

Now, combining like terms and reaaranging terms, we get

\(3x^{2}+4x+10\)

Thus, plugging in the numbers for the descriminant, we get

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

So -104 is our answer. 

Thanks! :)

Simplify as

3x^2 - kx + 33 = 0

This will have no real solutions when the discriminant is < 0

So

k^2  - 4(3) (33) < 0

k^2  - 396 < 0

k^2 < 396

k < sqrt (396)

k = 19

a + ab =   250

a -  ab =  -240          add these

2a = 10

a = 5

5 + 5b = 250

5b = 245

b = 245 / 5  =   49

(a, b)  = (5 , 49)

191 is prime

phi (191)  = 191 - 1  =  190 

1200 =  12 * 5^2 * 2^2  = 2^4 * 3 * 5^2

Using Euler's Totient Function

phi (1200)  =    1200 / ( 2 * 3 * 5) *  (2-1)(3-1) (5-1) = 40 (1)(2)(4)  =  320

Let's first take a look at the first equation. We want to put x in terms of y. 

So, solving the equation, we have

\(x+y=10-x-y\\ x=5-y\)

Now, subsituting this into the second equation, we get that

\(5-y-y=\left(5-y-1\right)+y-1\\\)

This simplifies to 

\(5-2y=3\)

Now, for this equation, we get that y is equal to 1. 

Thus, subsituting this value into x, we find that

\(x=5-1=4\)

Thus, the answers for the equation are

\(x=4,\:y=1\)

So our final answer is \((4, 1)\)

Thanks! :)

The volume of the solid is 24*pi.

First, let's note that 1/17 is a repeating decimal. 

We have that \(1/17 = 0.\overline{0588235294117647}\)

There are 16 repeating digits. 

Thus, we do 4000/16. We get

\(4000/16=250\)

Since it's divisible, it means the last digit of the repeating part is our answer, which is 7. 

So 7 is our answer. 

Thanks! :)

The slope of line segment PQ is 2 - sqrt(2).

phi(191) = 188.

phi(1200) = 1050

The 4000th digit is 2.

Solving the system, we get (x,y) = (3,2).

First, let's complete the square for both x and y to form the equation of a circle. We get

\(\left(x-5\right)^2+\left(y-\left(-4\right)\right)^2=\left(4\sqrt{3}\right)^2\)

This givies us a cricle with center of (5, -4) and a radius of \(4\sqrt3\)

Therefore, the area, which can be written as \(\pi r^2\), we have

\((4\sqrt3)^2 \pi = 16(3) \pi = 48 \pi \)

So 48pi is our amswer. 

Thanks! :)

There are 128 ways.

The answer is 3.

The number of ways is 520.

We can rewrite the given equation as follows:

\begin{align*} x^2+y^2 &= 2x-6y+6+8x-2y+1 \ (x^2+8x) + (y^2-8y) &= 7+6+1 \ (x^2+8x+16) + (y^2-8y+16) &= 30 \ (x+4)^{2} + (y-4)^{2} &= 30 = 5^2. \end{align*}

Thus, the equation represents a circle centered at (−4,4) with radius 5. Therefore, the area of the enclosed region is πr^2 = π(5^2) = 25π​.