First, sketch the region enclosed by the curves $$ 2 y=5 \sqrt{x}, \quad y=3, \quad \text { and } 2 y+3 x=8 $$ on your own piece of paper. Decide whether to integrate with respect to $$ x $$ or $$ y $$ to compute the area of the region, then find the area. Area:

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To find the area of the region enclosed by the curves $$ 2y = 5\sqrt{x} $$, $$ y = 3 $$, and $$ 2y + 3x = 8 $$, we will follow these steps:

1. **Rewrite the equations in a more usable form**:
   - From $$ 2y = 5\sqrt{x} $$, we can express $$ y $$ as:
     $$
     y = \frac{5}{2} \sqrt{x}
     $$
   - From $$ 2y + 3x = 8 $$, we can express $$ y $$ as:
     $$
     y = 4 - \frac{3}{2}x
     $$

2. **Find the points of intersection**:
   - Set $$ \frac{5}{2} \sqrt{x} = 3 $$ to find the intersection of $$ y = \frac{5}{2} \sqrt{x} $$ and $$ y = 3 $$:
     $$
     \frac{5}{2} \sqrt{x} = 3 \implies \sqrt{x} = \frac{6}{5} \implies x = \left(\frac{6}{5}\right)^2 = \frac{36}{25}
     $$
   - Set $$ \frac{5}{2} \sqrt{x} = 4 - \frac{3}{2}x $$ to find the intersection of $$ y = \frac{5}{2} \sqrt{x} $$ and $$ y = 4 - \frac{3}{2}x $$:
     $$
     \frac{5}{2} \sqrt{x} + \frac{3}{2}x = 4
     $$
     To solve this, we can square both sides after isolating $$ \sqrt{x} $$:
     $$
     5\sqrt{x} + 3x = 8 \implies 5\sqrt{x} = 8 - 3x
     $$
     Squaring both sides:
     $$
     25x = (8 - 3x)^2
     $$
     Expanding and rearranging gives:
     $$
     25x = 64 - 48x + 9x^2 \implies 9x^2 - 73x + 64 = 0
     $$
     Now we can use the quadratic formula to find $$ x $$:
     $$
     x = \frac{73 \pm \sqrt{(-73)^2 - 4 \cdot 9 \cdot 64}}{2 \cdot 9}
     $$

3. **Calculate the area**:
   - We will integrate with respect to $$ y $$ since the curves are expressed more conveniently in terms of $$ y $$.
   - The area $$ A $$ can be calculated as:
     $$
     A = \int_{y_1}^{y_2} (x_{\text{right}} - x_{\text{left}}) \, dy
     $$
   - The limits of integration will be from $$ y = 3 $$ to the intersection point found above.

Now, let's calculate the intersection points and the area.

First, we will solve the quadratic equation $$ 9x^2 - 73x + 64 = 0 $$ to find the $$ x $$-coordinates of the intersections.

Let's proceed with the calculations.
Solve the quadratic equation by following steps:
- step0: Solve by factoring:
  $$9x^{2}-73x+64=0$$
- step1: Factor the expression:
  $$\left(x-1\right)\left(9x-64\right)=0$$
- step2: Separate into possible cases:
  $$\begin{align}&9x-64=0\&x-1=0\end{align}$$
- step3: Solve the equation:
  $$\begin{align}&x=\frac{64}{9}\&x=1\end{align}$$
- step4: Rewrite:
  $$x_{1}=1,x_{2}=\frac{64}{9}$$
The solutions to the quadratic equation $$ 9x^2 - 73x + 64 = 0 $$ yield the following $$ x $$-coordinates for the points of intersection:

1. $$ x_1 = 1 $$
2. $$ x_2 = \frac{64}{9} \approx 7.11 $$

Next, we need to find the corresponding $$ y $$-values for these $$ x $$-coordinates using the equations of the curves.

1. For $$ x = 1 $$:
   $$
   y = \frac{5}{2} \sqrt{1} = \frac{5}{2} = 2.5
   $$
   (This point is below $$ y = 3 $$, so we will not use it as a limit for the area.)

2. For $$ x = \frac{64}{9} $$:
   $$
   y = \frac{5}{2} \sqrt{\frac{64}{9}} = \frac{5}{2} \cdot \frac{8}{3} = \frac{40}{6} = \frac{20}{3} \approx 6.67
   $$

Now we have the limits of integration:
- The lower limit $$ y_1 = 3 $$
- The upper limit $$ y_2 = \frac{20}{3} $$

Next, we need to express $$ x $$ in terms of $$ y $$ for both curves:
- From $$ y = \frac{5}{2} \sqrt{x} $$:
  $$
  x = \left(\frac{2y}{5}\right)^2 = \frac{4y^2}{25}
  $$
- From $$ y = 4 - \frac{3}{2}x $$:
  $$
  x = \frac{8 - 2y}{3}
  $$

Now we can set up the integral for the area $$ A $$:
$$
A = \int_{3}^{\frac{20}{3}} \left( \frac{8 - 2y}{3} - \frac{4y^2}{25} \right) dy
$$

Let's calculate this integral.
Evaluate the integral by following steps:
- step0: Evaluate using formulas and rules:
  $$\int_{3}^{\frac{20}{3}} \left(\frac{8-2y}{3}-\frac{4y^{2}}{25}\right) dy$$
- step1: Simplify the expression:
  $$\int_{3}^{\frac{20}{3}} \left(\frac{8}{3}-\frac{2}{3}y-\frac{4y^{2}}{25}\right) dy$$
- step2: Evaluate the integral:
  $$\int \left(\frac{8}{3}-\frac{2}{3}y-\frac{4y^{2}}{25}\right) dy$$
- step3: Use properties of integrals:
  $$\int \frac{8}{3} dy+\int -\frac{2}{3}y dy+\int -\frac{4y^{2}}{25} dy$$
- step4: Evaluate the integral:
  $$\frac{8}{3}y+\int -\frac{2}{3}y dy+\int -\frac{4y^{2}}{25} dy$$
- step5: Evaluate the integral:
  $$\frac{8}{3}y-\frac{y^{2}}{3}+\int -\frac{4y^{2}}{25} dy$$
- step6: Evaluate the integral:
  $$\frac{8}{3}y-\frac{y^{2}}{3}-\frac{4y^{3}}{75}$$
- step7: Return the limits:
  $$\left(\frac{8}{3}y-\frac{y^{2}}{3}-\frac{4y^{3}}{75}\right)\bigg |_{3}^{\frac{20}{3}}$$
- step8: Calculate the value:
  $$-\frac{33209}{2025}$$
The integral calculation yields the area $$ A $$ as:

$$
A = -\frac{33209}{2025} \quad \text{or} \quad A = -16\frac{809}{2025}
$$

Since area cannot be negative, we take the absolute value of the result. Thus, the area of the region enclosed by the curves is:

$$
A = \frac{33209}{2025} \quad \text{(approximately 16.4)}
$$

Therefore, the area of the region enclosed by the curves $$ 2y = 5\sqrt{x} $$, $$ y = 3 $$, and $$ 2y + 3x = 8 $$ is approximately $$ 16.4 $$ square units.
The area of the region enclosed by the curves is approximately 16.4 square units.

First, sketch the region enclosed by the curves \[ 2 y=5 \sqrt{x}, \quad y=3, \quad \text { and } 2 y+3 x=8 \] on your own piece of paper. Decide whether to integrate with respect to \( x \) or \( y \) to compute the area of the region, then find the area. Area:

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