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Solve por \( x \) n) \( b c=\frac{x^{2}-b^{2}-c^{2}}{2 \cos x} \) i) \( A=p(1+x)^{n} \) i) \( m n=p n-x \) k) \( y=a x^{2} \) 1) \( y=m x+c \)

Ask by Klein Young. in South Africa
Feb 20,2025

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Answer

Here are the solutions for each equation with respect to \( x \): 1. \( bc = \frac{x^{2} - b^{2} - c^{2}}{2 \cos x} \) No solution found for \( x \). 2. \( A = p(1+x)^{n} \) \( x = \left(\frac{A}{p}\right)^{\frac{1}{n}} - 1 \) 3. \( mn = pn - x \) \( x = -mn + np \) 4. \( y = ax^{2} \) \( x = \frac{\sqrt{ay}}{a} \) or \( x = -\frac{\sqrt{ay}}{a} \) 5. \( y = mx + c \) \( x = \frac{y - c}{m} \)

Solution

Solve the equation by following steps: - step0: Solve for \(x\): \(A=p\left(1+x\right)^{n}\) - step1: Swap the sides: \(p\left(1+x\right)^{n}=A\) - step2: Divide both sides: \(\frac{p\left(1+x\right)^{n}}{p}=\frac{A}{p}\) - step3: Divide the numbers: \(\left(1+x\right)^{n}=\frac{A}{p}\) - step4: Calculate: \(1+x=\left(\frac{A}{p}\right)^{\frac{1}{n}}\) - step5: Move the constant to the right side: \(x=\left(\frac{A}{p}\right)^{\frac{1}{n}}-1\) Solve the equation \( y=mx+c \). Solve the equation by following steps: - step0: Solve for \(x\): \(y=mx+c\) - step1: Swap the sides: \(mx+c=y\) - step2: Move the expression to the right side: \(mx=y-c\) - step3: Divide both sides: \(\frac{mx}{m}=\frac{y-c}{m}\) - step4: Divide the numbers: \(x=\frac{y-c}{m}\) Solve the equation \( mn=pn-x \). Solve the equation by following steps: - step0: Solve for \(x\): \(mn=pn-x\) - step1: Swap the sides: \(pn-x=mn\) - step2: Move the expression to the right side: \(-x=mn-pn\) - step3: Change the signs: \(x=-mn+pn\) - step4: Simplify: \(x=-mn+np\) Solve the equation \( y=ax^{2} \). Solve the equation by following steps: - step0: Solve for \(x\): \(y=ax^{2}\) - step1: Swap the sides: \(ax^{2}=y\) - step2: Divide both sides: \(\frac{ax^{2}}{a}=\frac{y}{a}\) - step3: Divide the numbers: \(x^{2}=\frac{y}{a}\) - step4: Simplify the expression: \(x=\pm \sqrt{\frac{y}{a}}\) - step5: Simplify: \(x=\pm \frac{\sqrt{ay}}{a}\) - step6: Separate into possible cases: \(\begin{align}&x=\frac{\sqrt{ay}}{a}\\&x=-\frac{\sqrt{ay}}{a}\end{align}\) Here are the solutions for each equation with respect to \( x \): 1. **Equation:** \( bc = \frac{x^{2} - b^{2} - c^{2}}{2 \cos x} \) **Solution:** No solution found for \( x \). 2. **Equation:** \( A = p(1+x)^{n} \) **Solution:** \[ x = \left(\frac{A}{p}\right)^{\frac{1}{n}} - 1 \] 3. **Equation:** \( mn = pn - x \) **Solution:** \[ x = -mn + np \] 4. **Equation:** \( y = ax^{2} \) **Solution:** \[ x = \frac{\sqrt{ay}}{a} \quad \text{or} \quad x = -\frac{\sqrt{ay}}{a} \] 5. **Equation:** \( y = mx + c \) **Solution:** \[ x = \frac{y - c}{m} \] If you need further assistance or explanations on any of these solutions, feel free to ask!

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Bonus Knowledge

¡Vamos a sumergirnos en resolver ecuaciones! Imaginemos que tienes que despejar \( x \) en todas estas interesantes fórmulas. Para cada caso, deberías aplicar diferentes estrategias, desde el uso de la trigonometría en la primera ecuación hasta manipular términos en el resto. Cada una de estas fórmulas es un pequeño puzzle que, al resolver, te da una nueva comprensión de relaciones matemáticas. Por ejemplo, en la ecuación \( y = ax^2 \), eso representa una parábola, ¡se convierte en una gráfica divertida! Puedes despejar \( x \) si tomas la raíz cuadrada de ambos lados, teniendo en cuenta que hay dos soluciones posibles (¡positivo y negativo!). En cada caso, asegúrate de analizar los valores posibles para \( x \) y sus implicaciones en el contexto dado; es como jugar un juego de estrategia con cada ecuación como una carta en tu mano.

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