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call the distance in miles "d". Call the time in minutes "t".  speed = d/t

40 mph = d/t

55 mph = d/(t-25)

can you solve the 2 equation 2 variable? 

Then the answer will be the d, because it is asking for the distance.

You can refer to the above question as a hint, but ill leave the solution below.

consider the psychology books as one unit. so the total ways is 5!/4!, 5! because there are 5 different units, and 4! because the math books are indistinguishable. (Im assuming), so the answer is 5

You can count the 4 freshmans as one group, call it x. so the total ways to arrange 3 sophomores and x is 4! which is 24. Multiply by 4! because the freshmans can stand in different orders next to each other, 4! different ways. so 4!x4!=576

The answer is 2/5.

The answer is f(n) = 6n^4 - 4n^2 + 12.

To ensure that \( f \) is its own inverse, we need \( f(f(x)) = x \) for all \( x \). Given the piecewise definition of \( f(x) \):

\[
f(x) =
\begin{cases}
k(x) & \text{if } x > 3 \\
x^2 - 6x + 12 & \text{if } x \leq 3
\end{cases}
\]

Let's first analyze the case when \( x \leq 3 \).

### Case 1: \( x \leq 3 \)

\[
f(x) = x^2 - 6x + 12
\]

We need to calculate \( f(f(x)) \):

1. Compute \( f(x) \):

\[
f(x) = x^2 - 6x + 12
\]

2. Find \( f(f(x)) \):

We substitute \( f(x) \) back into the function:

\[
f(f(x)) = f(x^2 - 6x + 12)
\]

Now we need to check whether \( x^2 - 6x + 12 \) is greater than 3 for \( x \leq 3 \). We can evaluate this quadratic:

First, calculate the value when \( x = 3 \):

\[
f(3) = 3^2 - 6 \cdot 3 + 12 = 9 - 18 + 12 = 3
\]

For \( x < 3 \), we check the maximum value of \( x^2 - 6x + 12 \):

The expression \( x^2 - 6x + 12 \) is always minimized at \( x = 3 \), and since it is a quadratic opening upwards, \( x^2 - 6x + 12 \) will always be greater than or equal to \( 3 \).

Thus, for \( x < 3 \):
- \( f(x) = x^2 - 6x + 12 \) produces a value \( y = x^2 - 6x + 12 > 3 \)

Thus \( f(f(x)) \) will use the case \( k(y) \).

3. Given \( y = x^2 - 6x + 12 \):

We have:

\[
k(y) = k(x^2 - 6x + 12)
\]

And we want this to equal \( x \):

\[
k(x^2 - 6x + 12) = x
\]

### Case 2: Define \( k(y) \)

We already know that for \( k(y) \) where \( y = x^2 - 6x + 12 \):

We need \( k(y) \) when \( y > 3 \).

To find a suitable \( k(y) \), consider a simple equation with a variable transformation. Suppose we let \( k(y) = g(y) \), where we define \( g(y) \) such that \( f(g(y)) = y \).

Using the quadratic equation:

If \( y > 3 \):

Let \( k(y) \) be derived from \( k(x) \):

We need the outcome to solve \( g(y) = \sqrt{y} - 3\):

### Resulting function

Assuming \( y = k(x) \):

Then,

\[
k(y) = -3 + \sqrt{y - 3}
\]

### Summary of \( k(x) \)

Thus, we can derive:

\[
k(x) = -3 + \sqrt{x - 3} \quad \text{for } x > 3
\]

Thus, the function \( f(x) \) such that \( f \) is its own inverse is:

\[
f(x) =
\begin{cases}
-3 + \sqrt{x - 3} & \text{if } x > 3 \\
x^2 - 6x + 12 & \text{if } x \leq 3
\end{cases}
\]

Confirming:

- \( k(3) \) returns back confirming f is its own inverse,
- Further validations can ensure continuality and completeness over all functional parts.

Therefore, the final expression for \( k(x) \) is:

\[
\boxed{-3 + \sqrt{x - 3}} \quad \text{for } x > 3
\]

To solve the problem, we will first calculate the total number of ways to paint the cube and then determine the number of ways to paint the cube such that no two adjacent (sharing an edge) faces are the same color. Finally, we will use these counts to find the desired probability.

1. **Total ways to paint the cube:**

Each of the 6 faces of the cube can be painted in one of 6 colors (red, orange, yellow, green, blue, purple). Therefore, the total number of ways to paint all the faces of the cube is given by:

\[
6^6 = 46656
\]

2. **Ways to paint the cube with no adjacent faces the same color:**

We will use the principle of complementary counting. Let’s find how many ways there are to paint the cube such that no two adjacent faces share the same color.

We will use the chromatic polynomial for a cube which can be represented as a graph. The graph of a cube has 8 vertices (the corners of the cube) and 12 edges (the edges of the cube). Each face of the cube is adjacent to 4 other faces.

To find the number of valid colorings, we can use Kő’s theorem which provides that the number of ways to color the graph with \( k \) colors such that no two adjacent vertices share the same color is given by the chromatic polynomial \( P(G, k) \). For a cube, the formula is:

\[
P(G, k) = k(k - 1)^2(k - 2)^3
\]

Here, \( k \) is the number of colors available. Plugging in our \( k = 6 \):

\[
P(G, 6) = 6 \cdot 5^2 \cdot 4^3
\]

Calculating this step by step:

- \( 5^2 = 25 \)
- \( 4^3 = 64 \)

So, we compute:

\[
P(G, 6) = 6 \cdot 25 \cdot 64
\]

Calculating this gives:

\[
6 \cdot 25 = 150
\]
\[
150 \cdot 64 = 9600
\]

Thus, the total number of ways to paint the cube such that no two adjacent faces are the same color is:

\[
9600
\]

3. **Calculating the number of ways to paint the cube such that at least one pair of adjacent faces is the same color:**

We now will use complementary counting to find the number of ways where at least one pair of adjacent faces are the same color. This is simply the difference between the total number of paintings and the paintings with no adjacent faces being the same color:

\[
\text{Unwanted cases} = 6^6 - P(G, 6) = 46656 - 9600 = 37056
\]

4. **Finding the probability:**

The probability that the cube has at least one pair of adjacent faces sharing the same color is given by the ratio of the unwanted cases to the total cases:

\[
P(\text{at least one pair same}) = \frac{37056}{46656}
\]

Calculating this,

\[
P = \frac{37056}{46656} \approx 0.7935
\]

Finally, rounding to the nearest thousandth gives us:

\[
\boxed{0.794}
\]

When Dirk finishes, Edith has run 14km  and Foley has run 9 km

This means that Foley runs 9/14 as fast as Edith

When Edith has run 24km, Foley has run \(24 (9/14) ≈ 15.43 km\)

So Foley is \(24 - 15.43 ≈ 8.57\) km behind!

Thanks! :)

Find slope    m = (8-17) / (5- -4) = -9/9 = -1 

y = mx + b    then becomes 

 y = -x + b    sub ineither of the points to solve for b = 13 

y = -x + 13 

x+y = 13 

1/13 x  +  1/13  y  = 1          a = 1/13 

As posted before:

(n+2) / n    *   ( n+3)/ (n+1)  *  (n+4) / (n+2) = 10      <==== Given 

Simplifies to 

1/n * (n+3) / ((n+1) * (n+4) )= 10 

then:

(n^2 + 7n + 12 )/ (n^2 +n) = 10        Cross multiply and simplify to get :

9n^2 + 3n-12 = 0 

Use Quadratic Formula with    a = 9    b= 3    and c = -12     to find  n = 1      (or - 4/3  <=== does not work) 

So the first fraction would be 3/1            

                                                    x^2 - 5x + 4 = -x^2 - 25x - 14   

                                                    2x² + 20x + 18  =  0     

                                                      x² + 10x + 9  =  0    

                                                      (x + 1)(x + 9)  =  0    

                                                                 (x + 1) = 0     ==>>     x = –1    

                                                                 (x + 9) = 0     ==>>     x = –9    

                                                                  x = –1, –9    

_.   

The length of AB is 10*sqrt(3).

The area of WXYZ is 15.

To solve for the coordinates of point \( A \), let’s first analyze the properties of the given parabola and the geometry involved.

The equation of the parabola is given by

\[
x^2 = 4y.
\]

The focus \( F \) of this parabola is at \( (0, 1) \). We can derive the directrix of the parabola from its standard form, which implies:

- The distance from any point \( (x, y) \) on the parabola to the focus is equal to its distance to the directrix line \( y = -1 \).

Next, let’s consider a line passing through the focus \( F \) with a slope \( m \):

\[
y - 1 = m(x - 0) \quad \Rightarrow \quad y = mx + 1.
\]

We now substitute this expression for \( y \) into the parabola equation to find the intersection points \( M \) and \( N \).

Substituting \( y = mx + 1 \) into \( x^2 = 4y \), we get:

\[
x^2 = 4(mx + 1).
\]

This results in the quadratic equation:

\[
x^2 - 4mx - 4 = 0.
\]

Using the quadratic formula, we can find the \( x \)-coordinates of points \( M \) and \( N \):

\[
x = \frac{4m \pm \sqrt{(4m)^2 + 16}}{2} = 2m \pm 2\sqrt{m^2 + 1}.
\]

This corresponds to \( x_M = 2m + 2\sqrt{m^2 + 1} \) and \( x_N = 2m - 2\sqrt{m^2 + 1} \).

Next, we can determine the corresponding \( y \)-coordinates of points \( M \) and \( N \):

For \( M \):

\[
y_M = m(2m + 2\sqrt{m^2 + 1}) + 1 = 2m^2 + 2m\sqrt{m^2 + 1} + 1,
\]

For \( N \):

\[
y_N = m(2m - 2\sqrt{m^2 + 1}) + 1 = 2m^2 - 2m\sqrt{m^2 + 1} + 1.
\]

Now, we want to find \( A \) such that angle \( MAN = 90^\circ \), meaning line \( FA \) is perpendicular to line \( MN \).

To do this, we firstly determine the slope \( m_{MN} \) of line segment \( MN \):

\[
m_{MN} = \frac{y_M - y_N}{x_M - x_N} = \frac{(2m^2 + 2m\sqrt{m^2 + 1} + 1) - (2m^2 - 2m\sqrt{m^2 + 1} + 1)}{(2m + 2\sqrt{m^2 + 1}) - (2m - 2\sqrt{m^2 + 1})}
\]

This simplifies to:

\[
m_{MN} = \frac{(2m + 2\sqrt{m^2 + 1} - (2m - 2\sqrt{m^2 + 1}))}{4\sqrt{m^2 + 1}} = \frac{4\sqrt{m^2 + 1}}{4\sqrt{m^2 + 1}} = 1.
\]

Consequently, the slope \( m_{FA} \) of line \( FA \) must be \( -1 \) (since \( m_{FA} \cdot m_{MN} = -1 \)):

\[
y - 1 = -1(x - 0) \quad \Rightarrow \quad y = -x + 1.
\]

We must find the intersection of this line \( y = -x + 1 \) with the parabola \( x^2 = 4y \).

Substituting \( y = -x + 1 \) into the parabola’s equation gives:

\[
x^2 = 4(-x + 1).
\]

Rearranging leads to:

\[
x^2 + 4x - 4 = 0.
\]

Applying the quadratic formula yields:

\[
x = \frac{-4 \pm \sqrt{4^2 + 4 \cdot 4}}{2} = \frac{-4 \pm \sqrt{16 + 16}}{2} = \frac{-4 \pm \sqrt{32}}{2} = -2 \pm 2\sqrt{2}.
\]

Taking the positive root in the first quadrant:

\[
x_A = -2 + 2\sqrt{2}.
\]

Then utilizing \( y = -x + 1 \):

\[
y_A = -(-2 + 2\sqrt{2}) + 1 = 2 - 2\sqrt{2} + 1 = 3 - 2\sqrt{2}.
\]

Thus, the coordinates of point \( A \) are

\[
\boxed{(-2 + 2\sqrt{2}, 3 - 2\sqrt{2})}.
\]

We can use the information we already know to solve this problem!

First off, let's note that \((x+y)^3 = x^{3}+3x^{2}y+3xy^{2}+y^{3}\), which we can simplify to \(8=5+6xy\). Solving for xy, we get \(xy=1/2\). 

This may seem useless, but it will come into play later. 

Now, we know that \((x+y)^2 = x^2+2xy+y^2\), which contains x^2+y^2. Isolating this, we get \(x^2+y^2 = (x+y)^2 - 2xy\). 

We already know all the terms! We can plug in 1/2 and 2 which we know from above! We get \(x^2 + y^2 = 4 - 1 = 3\). 

So 3 is our answer!

Thanks! :)