gpt-oss-13.1b-specialized-math-pruned-moe-only-19-experts

<|channel|>analysis<|message|>We need to solve linear equation. Provide step-by-step.<|end|><|start|>assistant<|channel|>final<|message|>**Equation to solve**

\[
2x + 5 = 17
\]

---

### Step 1: Isolate the term containing \(x\)

Subtract \(5\) from both sides of the equation to get the \(x\)-term alone on the left side.

\[
\begin{aligned}
2x + 5 &= 17 \\
2x + 5 - 5 &= 17 - 5 \\
2x &= 12
\endaligned
\]

---

### Step 2: Solve for \(x\)

Divide both sides by the coefficient of \(x\), which is \(2\).

\[
\beginaligned
\frac{2x}{2} &= \frac{12}{2} \\
x &= 6
\endaligned
\]

---

### Step 3: Verify the solution

Plug \(x = 6\) back into the original equation:

\[
2(6) + 5 = 12 + 5 = 17
\]

The left side equals the right side, so the solution is correct.

---

\[
\boxed{x = 6}
\]<|return|>

<|channel|>analysis<|message|>Need explain theorem, formula, usage.<|end|><|start|>assistant<|channel|>final<|message|>**The Pythagorean Theorem**

> In a right‑angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) equals the sum of the squares of the lengths of the other two sides.

Mathematically:

\[
a^{2} + b^{2} = c^{2}
\]

where  
* \(a\) and \(b\) are the legs (the two sides that form the right angle),  
* \(c\) is the hypotenuse.

---

## How it’s used

| **Scenario** | **What you need** | **How the theorem helps** |
|--------------|-------------------|---------------------------|
| **Finding a missing side** | One leg and the hypotenuse, or both legs | Rearsolve the equation:  
\(c = \sqrt{a^{2}+b^{2}}\) or \(a = \sqrt{c^{2}-b^{2}}\) |
| **Checking if a triangle is right‑angled** | All three side lengths | Verify if \(a^{2}+b^{2}=c^{2}\) (within rounding error) |
| **Geometry & construction** | Designing right‑angled shapes, ladders, bridges | Ensures structural angles are 90° |
| **Trigonometry** | Relating sine, cosine, and tangent | \( \sin^{2}\theta + \cos^{2}\theta = 1\) comes from the theorem |
| **Distance in coordinate geometry** | Distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) | \(d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\) |
| **Physics** | Projectile motion, velocity vectors, etc. | Decompose vectors into perpendicular components |

---

### Quick Examples

1. **Right triangle with legs 3 m and 4 m**  
   \(c = \sqrt{3^{2}+4^{2}} = \sqrt{9+16} = \sqrt{25} = 5\) m.

2. **Triangle with hypotenuse 10 m and one leg 6 m**  
   \(b = \sqrt{10^{2}-6^{2}} = \sqrt{100-36} =

<|channel|>analysis<|message|>Need area = πr^2 = π*49 ≈ 153.938.<|end|><|start|>assistant<|channel|>final<|message|>The area \(A\) of a circle is given by

\[
A = \pi r^{2}
\]

where \(r\) is the radius.

For a radius of \(7\) m:

\[
A = \pi (7\ \text{m})^{2}
   = \pi \times 49\ \text{m}^{2}
   \approx 3.14159 \times 49\ \text{m}^{2}
   \approx 153.938\ \text{m}^{2}
\]

So the circle has an area of about **\(154\ \text{m}^{2}\)** (rounded to the nearest square meter).<|return|>