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What is the fifth smallest prime number greater than 100?    

The first five primes greater than 100 are 101, 103, 107, 109, and 113.   

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Find the constant k such that the quadratic 2x^2 + 3x + 8x - x^2 + 4x + k has a double root.   

                                        2x² + 3x + 8x – x² + 4x + k   

Combine like terms           x² + 15x + k    

In math, the term "double root" means "repeated root" ... i.e., both roots are the same   

Examples:    x² + 2x + 1    factors to (x+1)(x+1) and both roots are –1   

                    x² + 6x + 9    factors to (x+3)(x+3) and both roots are –3   

                    x² – 8x + 16  factors to (x–4)(x–4) and both roots are +4   

One way to determine that a quadratic has a double root is when its discriminant equals zero   

The discriminant is the expression underneath the radical in the quadratic equation   

When a quadratic is posed as ax² + bx + c the discriminant is b² – 4ac   

In x² + 15x +k in this problem, the discriminant is 15² – (4)(1)(k)   

Multiply it out, set equal to 0, and solve for k       225 – 4k = 0   

                                                                                    –4k  =  –225   

                                                                                        k  =  56.25    

The problem doesn't ask for the repeated root, but it's the square root of k.    

The factors of the quadratic are (x + 7.5)(x + 7.5) and the repeated root is –7.5   

check answer   

                                      x² + 15x + 56.25 = 0   

                                     (–7.5)² + (15)(–7.5) + 56.25 = 0   

                                       56.25 – 112.5 + 56.25 = 0    

                                                                       0 = 0   

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The parabola y = ax^2 + bx + c is graphed below. Find a + b + c. (The grid lines are one unit apart.)    
The parabola passes through (-2,8), (0,0), and (2,8).   

There is no graph shown.  Relying only on the three points provided,    

one parabola that passes through (–2,8), (0,0), and (2,8) is y = 2x²    

If that is correct, then a + b + c  =  2 + 0 + 0  =  2.   

Those three points, and the vertex at (0,0), indicate that the parabola    

is symmetrical to the y-axis and that the equation has a double root.  

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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! :)

a + b = 1    square both sides

a^2 + b^2 + 2ab = 1

     2  + 2ab = 1

2ab = -1

ab = -1/2

a^3 + b^3  =  (a + b) ( a^2 + b^2 -ab) = (1) (2 - -1/2) = 2 + 1/2  =  5/2

a11 - a10 = a9

1 - 4 = -3 

a10 - a9  = a8

4  - - 3  = 7

a9 - a8 = a7

-3 - 7  = -10

a8 - a7 = a6

7 - -10 = 17

Oh, my stars.  I misread the problem.  I thought it said x² – mx + 24 = 0,  

CPhill solved it correctly at https://web2.0calc.com/questions/help-help_55   

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Simplify as

x^2 -mx + 14 =  0

Note that  the possible factorizations are

( x - 7) (x - 2)      m =  9

( x + 7) (x + 2)    m = -9

(x - 14) ( x -1)    m = 14

(x + 14) ( x + 1)  m = -14

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! :)

a^3 - 1/a^3  =   (a -1/a) (a^2 + 1 + 1/a^2) =  (0) ( a^2 + 1 + 1/b^2)  =  0

Let the roots be a,b

Product of the roots  = ab =  -4   →  2ab = -8

Sum of the roots =  a + b = -12     square both sides

a^2 + b^2 + 2ab = 144

a^2 + b^2 - 8 =144

a^2 + b^2 =  152

A

                X

                     5

B                        C

            15

Since BXA = 90  then  BXC also = 90

So BXC is aright triangle such that

BC^2 - CX^2  = BX^2

15^2 - 5^2  = BX^2

200 = BX^2

BX  = sqrt 200  =   10sqrt 2  ≈  14.14

a + b = 4      square both sides

a^2 + 2ab + b^2  =16

a^2 + b^2 + 2ab = 16

6   + 2ab =  16

2ab =  10

ab = 5

a^3 + b^3  = (a + b) (a^2 + b^2  - ab) =  (4) (6 - 5)  =   4

                      A

               30       30

B                      D                        C

Impossible

If AD is a bisector, then CAD  =  BAC / 2 =   30

So CAD cannot = 45

We can use Heron's formula to first find the area of the triangle, \(\sqrt{s(s-a)(s-b)(s-c)}\). Since the semiperimeter is \(17\), we have 

\(\sqrt{17(17-15)(17-9)(17-10)} = \sqrt{1904} = 4\sqrt{119}\)

as the area of triangle \(ABC\). The shortest altitude of a triangle has the longest side as its base, so we have 

\(\frac{15h}{2} = 4\sqrt{119}\),

where \(h\) is the altitude. Solving for \(h\) yields \(h = \boxed{\frac{8\sqrt{119}}{15}}\).

^{(I'm not sure if there's a way to solve this without using Heron's, but if there is, please let me know! Thanks.)}

Simplifying we get \(x^2 + 12x - 4 = 0\). Using the quadratic formula we find the roots to be \(-6 + 2\sqrt{10}\) and \(-6 - 2\sqrt{10}\). Taking their squares, we get \(36 + 4\sqrt{10} + 40\) and \(36 - 4\sqrt{10} + 40\). Adding them together, we get \(\boxed{152}\)

Since

    (1)   =  (1)       set the equations  = 

x^2 + (y-1)^2  = (x-1)^2 + y^2     simplify

x^2 + y^2 - 2y + 1 =  x^2 -2x + 1 + y^2

-2y = -2x

x = y

So

x^2 + ( x - 1)^2  =  1

x^2 + x^2 -2x + 1   = 1

2x^2 - 2x  =  0

x^2 - x = 0

x ( x -1) = 0

x = 0       x   = 1

y = 0      y = 1

The intersection points are   ( 0, 0)  and (1, 1)

The distance  between them =   sqrt [1^2 + 1^2]  = sqrt 2

Expanding \((x-2)(x+2)\) gives \(x^2 - 4\). In order to get it into the form \(Ax^2+Bx+1\) where the constant term is \(1\), we can divide the original quadratic by \(-4\) to get \(-\frac{1}{4}x^2 + 1\), so \(A+B = -\frac{1}{4} + 0 = \boxed{-\frac{1}{4}}\)

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! :)

Or did you mean to find the units digit of \(13^{287}\)? It's a bit more complicated but still doable.