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Here's how to calculate the minimal amount you should bid per bridge for this three-year project:

1. Consider the Equipment:

Equipment cost: $1,000,000

Salvage value (selling price after 3 years): $400,000

Depreciation: Straight-line over 3 years

Annual depreciation expense: ($1,000,000 - $400,000) / 3 years = $200,000/year

2. Factor in Working Capital:

Net working capital needed: $250,000 (constant throughout the project)

3. Analyze Costs and Taxes:

Fixed cost per year: $500,000

Variable cost per bridge: $3,000,000

Required rate of return: 20%

Tax rate: 10%

4. Calculate After-Tax Cash Flow per Year:

We'll consider one year at a time. Let's denote the year as "Y":

Revenue from the bridge (unknown yet - this is what we're solving for): "Bid amount (X)"

Total cost: Fixed cost + Variable cost - Depreciation expense

Taxable income before depreciation: Bid amount (X) - Total cost

Depreciation tax shield: Annual depreciation expense * Tax rate = $200,000 * 10% = $20,000

Taxable income: Taxable income before depreciation - Depreciation tax shield

Taxes (at 10% rate): Taxable income * Tax rate

After-tax cash flow (Y): Bid amount (X) - Total cost - Taxes + Depreciation expense + Net working capital

5. Apply Annuity Factor for Present Value:

We're considering a three-year project, so we need to find the present value of the after-tax cash flow for each year. Since the cash flow happens at the end of each year (annuity due), we'll use an annuity factor (considering the required rate of return of 20%).

Annuity factor for 3 years at 20% interest: 2.106 (given)

6. Solve for the Minimum Bid (X):

We want the project's Net Present Value (NPV) to be zero at the minimum acceptable bid. NPV is the sum of the present values of all future cash flows.

Net Present Value (NPV) Equation:

NPV = After-tax cash flow (Year 1) + After-tax cash flow (Year 2) + After-tax cash flow (Year 3) - Initial equipment cost + Salvage value = 0

We can rewrite the above equation with the annuity factor and solve for X (the minimum bid amount).

After rearranging and substituting the after-tax cash flow formula from step 4:

X * (1 - (1 + Required return)^(-Project life)) / Required return = (Total cost - Depreciation expense + Net working capital) * Annuity factor + Equipment cost - Salvage value

Plugging in the numbers:

X * (1 - (1 + 0.2)^(-3)) / 0.2 = ($500,000 + $3,000,000 - $200,000 + $250,000) * 2.106 + $1,000,000 - $400,000

X * (1 - 0.512) / 0.2 = $3,856,000 * 2.106 + $600,000

X * 0.488 = $8,038,16 + $600,000

X = ($8,038,16 + $600,000) / 0.488 ≈ $8,647,764

Minimum Bid amount rounded to nearest $100: $8,647,800

Therefore, the minimum amount you should bid per bridge to achieve a zero Net Present Value for this three-year project is approximately $8,647,800.

Let's rewrite the equation $n = ab + 3a + 7b$ as $n = a(b+3) + 7b$.

We can factor the equation as $n = (a+7)(b+3) - 21$.

Since $70 \leq n \leq 90$, we have $49 \leq (a+7)(b+3) \leq 77$.

Factoring the bounds, we get $49 = 7 \times 7$ and $77 = 11 \times 7$.

So the possible values for $(a+7)(b+3)$ are $49, 56, 63, 70, 77$.

The factors of these numbers are:

$49$: $(1, 49), (7, 7)$

$56$: $(1, 56), (2, 28), (4, 14), (7, 8)$

$63$: $(1, 63), (3, 21), (7, 9)$

$70$: $(1, 70), (2, 35), (5, 14), (7, 10)$

$77$: $(1, 77), (7, 11)$

Therefore, there are $\boxed{10}$ integers $n$ with $70 \leq n \leq 90$ that can be written as $n = ab + 3a + 7b$ for at least one ordered pair of positive integers $(a, b)$.

Let the point $(x, y)$ be $5$ units away from the point $(2, 7)$. This means that the distance between $(x, y)$ and $(2, 7)$ is $5$ units.

Using the distance formula, we have

\[\sqrt{(x - 2)^2 + (y - 7)^2} = 5.\]

Squaring both sides, we get

\[(x - 2)^2 + (y - 7)^2 = 25.\]

Since the point $(x, y)$ lies on the line $y = 5x - 28$, we substitute $y = 5x - 28$ into the equation above and solve for $x$:

\[(x - 2)^2 + (5x - 35)^2 = 25.\]

\[x^2 - 4x + 4 + 25x^2 - 350x + 1225 = 25.\]

\[26x^2 - 354x + 1205 = 0.\]

The solutions to this quadratic equation for $x$ are $x = 5$ and $x = \frac{23}{13}$. Plugging these values of $x$ back into the equation $y = 5x - 28$, we find that the corresponding $y$ values are $y = 2$ and $y = 1$, respectively.

Therefore, the points that are $5$ units away from $(2, 7)$ and lie on the line $y = 5x - 28$ are $(5, 2)$ and $\left(\frac{23}{13}, 1\right)$.

We need to find the length \( AB \).

First, let's use the Power of a Point theorem, which states that for a point \( X \) outside a circle, the product of the lengths of the segments of one secant line through \( X \) is equal to the square of the length of the tangent segment from \( X \) to the circle. That is:

\[
XY^2 = XA \cdot XB
\]

We know:

\[
XY = 8 \implies XY^2 = 8^2 = 64
\]

\[
XA = 20
\]

Let \( XB = x \). Then:

\[
XA \cdot XB = 20 \cdot x
\]

According to the Power of a Point theorem:

\[
20 \cdot x = 64
\]

Solving for \( x \):

\[
x = \frac{64}{20} = 3.2
\]

Therefore, the correct length of AB is 24.