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Let's consider two functions \(f(x)\) and \(g(x)\) that satisfy the following differential equations:

1. \(f'(x) = 2x\)  
2. \(g''(x) = -\sin(x)\)

Let's solve these differential equations and find the expressions for \(f(x)\) and \(g(x)\):

**1. Differential Equation for \(f(x)\):**

Given \(f'(x) = 2x\), we can integrate both sides to find \(f(x)\):

\[\int f'(x) \, dx = \int 2x \, dx\]

\[f(x) = x^2 + C_1\]

Here, \(C_1\) is the constant of integration.

**2. Differential Equation for \(g(x)\):**

Given \(g''(x) = -\sin(x)\), we can integrate twice to find \(g(x)\):

\[\int g''(x) \, dx = \int -\sin(x) \, dx\]

\[g'(x) = \cos(x) + C_2\]

Integrate again:

\[\int g'(x) \, dx = \int (\cos(x) + C_2) \, dx\]

\[g(x) = \sin(x) + C_2x + C_3\]

Here, \(C_2\) and \(C_3\) are constants of integration.

So, the expressions for \(f(x)\) and \(g(x)\) are:

\[f(x) = x^2 + C_1\]
\[g(x) = \sin(x) + C_2x + C_3\]

These expressions satisfy the given differential equations. Keep in mind that the constants of integration (\(C_1\), \(C_2\), and \(C_3\)) can take any real values, and the solutions may have different forms based on the specific values of these constants.

Let's denote the sides of the quadrilateral as \(a\), \(b\), \(c\), and \(d\). Given that the perimeter is 52, we have:

\[a + b + c + d = 52.\]

We're also given that the quadrilateral has an inscribed circle with a radius of 5. An inscribed circle touches the quadrilateral at its four sides, so the sum of the lengths of opposite sides of the quadrilateral is equal to its perimeter. That is:

\[a + c = b + d = 52/2 = 26.\]

We can use these equalities to express the sides of the quadrilateral in terms of \(a\) and \(b\):

\[c = 26 - a, \quad d = 26 - b.\]

The area of a quadrilateral inscribed in a circle can be calculated using the formula:

\[Area = \sqrt{(s - a)(s - b)(s - c)(s - d)},\]

where \(s\) is the semiperimeter (\(s = \frac{a + b + c + d}{2}\)).

Substituting the values we have:

\[s = \frac{a + b + c + d}{2} = \frac{52}{2} = 26.\]

\[Area = \sqrt{(26 - a)(26 - b)(26 - c)(26 - d)}.\]

Substituting the expressions for \(c\) and \(d\) in terms of \(a\) and \(b\):

\[Area = \sqrt{(26 - a)(26 - b)(a)(b)}.\]

We want to maximize the area, which occurs when \(a\) and \(b\) are equal. So, let's assume \(a = b\):

\[Area = \sqrt{(26 - a)(26 - a)(a)(a)} = \sqrt{(26 - a)^2 a^2} = (26 - a) a.\]

To maximize the area, we should choose \(a\) such that the expression \((26 - a) a\) is maximized. This occurs when \(a = 13\) (the midpoint between 0 and 26):

\[Area = (26 - 13) \cdot 13 = 13 \cdot 13 = 169.\]

So, the area of the quadrilateral is 169 square units.

Given that there are two circles with radii 2 and 6, and their centers are 10 units apart, we want to find the distance between the two tangency points as shown in the diagram.

Let's denote the centers of the circles as \(A\) and \(B\), with radii \(r_1 = 2\) and \(r_2 = 6\) respectively. The distance between the centers is \(AB = 10\).

We can form a right triangle using the centers of the circles and one of the tangency points, let's call it \(T\), as shown below:

      T
       |\
       | \
       |  \   6
     A |   \
       |    \
       |     \
       |______\
         10   B
 

Using the Pythagorean theorem, we can find the distance between the tangency point \(T\) and the center \(A\):

\[AT^2 = AB^2 - BT^2 = 10^2 - r_1^2 = 100 - 4 = 96.\]

Now, we can find the distance between the tangency point \(T\) and the center \(B\):

\[BT^2 = AB^2 - AT^2 = 10^2 - r_2^2 = 100 - 36 = 64.\]

Using the distance formula, we have \(BT = \sqrt{64} = 8\).

So, the distance between the two tangency points as shown in the diagram is \(2 \times 8 = 16\) units.

The triangle will be isosceles  with two equal legs

So

L^2 + L^2  =  6^2

2L^2 = 36

L^2  = 36/2

L^2   =18

Area =  (1/2)  (product of the  legs)  = (1/2) (L * L ) =  (1/2) (L^2) =  (1/2)(18)  = 9 in^2

Let's approach this step by step:

1. Draw the diagram with circles centered at A and B and connect the centers A and B to form the line segment AB of length 6.

2. Draw the tangent line that is common to both circles and intersects AB at a point, let's call it E. The point where the tangent line touches the smaller circle is C, and the point where it touches the larger circle is D.

3. You have two right triangles: ACE and BDE. The hypotenuses AC and BD are the radii of the circles, which are 1 and 2 respectively.

4. You can calculate the lengths of CE and DE using the Pythagorean theorem:
   - In triangle ACE: \(CE^2 = AC^2 - AE^2 = 1^2 - 4^2 = -15.\)
   - In triangle BDE: \(DE^2 = BD^2 - BE^2 = 2^2 - 4^2 = 0.\)

   Since CE squared is negative, it's not a real number, and DE squared is 0, which implies DE is 0. However, this doesn't make sense geometrically, so there seems to be an issue with the problem statement or the setup.

The situation described doesn't seem to align with the properties of the given circles and their common internal tangent. Double-check the information provided and the diagram to make sure everything is accurate and coherent.

Sub the first equation   into the  second  and we  have

y^4 + y^2 = 1 - y^4

2y^4 + y^2  - 1  =   0       factor this

(2y^2 -1) (y^2 + 1) = 0

The second factor has no real  solution......set the first factor  = 0

So

2y^2 - 1  =  0 

2y^2  = 1

y^2 =  1/2

y = 1/sqrt (2)     or  y = -1/sqrt (2)

And x =  y^4  =   (1 / sqrt (2))^4 =  1/4

The distance between  these points  is  just     1/sqrt (2)  -  -1/sqrt (2)  = 2/sqrt (2) =  sqrt (2) =  0 + 1sqrt (2)

(u,v)  = (0, 1)

minus 5x on both sides

2y=-5x-9

divide by 2

y=-5/2x-9/2

in format, y=mx+b

-5/2 is the m here

so

\(\frac 5 2\)

x=-73/25

y=1 6/25

z=1/10

SpectraSynth: 

You made a typo at this step:

14,400 / 8,000 ==1.8

1.8^(1/16) ==1.0374198 - 1 x 400 ==14.967919% - nominal annual rate compounded quarterly.

By Pitot Theorem, \(AB+CD = AC+BD = 12\), so \(CD=7\). 

Let's consider the properties of a quadrilateral with an inscribed circle (an incircle). One of the key properties is that the opposite sides of the quadrilateral are tangents to the incircle, and these tangents are equal in length.

Given that the perimeter of the quadrilateral ABCD is 24 and AB = 5, we can label the other sides as follows: BC = x, CD = y, and DA = z.

So, we have:

AB + BC + CD + DA = 24

Substituting the given values:

5 + x + y + z = 24

Solving for z:

z = 24 - 5 - x - y z = 19 - x - y

Since opposite sides of an inscribed quadrilateral are equal in length, we know that AD = BC and CD = AB.

So, we have:

AD = BC = x CD = AB = 5

Now, we can rewrite the perimeter equation with these lengths:

x + 5 + x + y + 19 - x - y = 24

Simplify:

2x + 24 = 24

Subtract 24 from both sides:

2x = 0

Divide by 2:

x = 0

However, a side length cannot be zero, which means there might be an issue with the problem statement or the approach taken. Please double-check the values and conditions given in the problem to ensure accuracy. If there is additional information or clarification, I would be happy to help further.

Here is a hint: use inverse modulo to get that \(n \equiv 3 \pmod5\)

LOL!!!!......can't argue with that !!!!