CALCULO I - CALCULO NIVELI 6- Cuando dos resistencias $$ R_{1} $$ y $$ R_{2} $$ están conectadas en paralelo, la resistencia total $$ R $$ satisface la ecuación $$ \frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}} $$. Encuentre $$ R_{1} $$ para un circuito en paralelo en el que $$ R_{2}=4[ $$ ohm $$ ] $$ y $$ R $$ debe ser al menos de $$ 2,5[ $$ ohm $$ ] $$. 7- Encuentre y grafique en la recta real el conjunto solución de $$ \begin{array}{lll} ext { a) }|4+7 x|=5 & ext { b) }|x-8|=\mid 5 x-16 & ext { e) }\left|x+\frac{2}{3} ight|=|-2 x+3| \ ext { d) } 23-|8 x|=7-x^{2} & ext { e) }|9-x| \leq \sqrt{5} & ext { f) } 12-|x+5|

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CALCULO I - CALCULO NIVELI 6- Cuando dos resistencias $$ R_{1} $$ y $$ R_{2} $$ están conectadas en paralelo, la resistencia total $$ R $$ satisface la ecuación $$ \frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}} $$. Encuentre $$ R_{1} $$ para un circuito en paralelo en el que $$ R_{2}=4[ $$ ohm $$ ] $$ y $$ R $$ debe ser al menos de $$ 2,5[ $$ ohm $$ ] $$. 7- Encuentre y grafique en la recta real el conjunto solución de $$ \begin{array}{lll}	ext { a) }|4+7 x|=5 & 	ext { b) }|x-8|=\mid 5 x-16 & 	ext { e) }\left|x+\frac{2}{3}ight|=|-2 x+3| \ 	ext { d) } 23-|8 x|=7-x^{2} & 	ext { e) }|9-x| \leq \sqrt{5} & 	ext { f) } 12-|x+5|<0 \ 	ext { g) } 49-7 x^{2} \leq 0 & 	ext { h) }-1+\left(x+\frac{5}{3}ight)^{2} \leq 2 & 	ext { i) } \frac{x+9}{|x-5| x^{5}+0}<\end{array} $$

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Solve the inequality by following steps: - step0: Solve for $$x$$: $$12-\left|x+5\right|<0$$ - step1: Rearrange the terms: $$-\left|x+5\right|<-12$$ - step2: Calculate: $$\left|x+5\right|>12$$ - step3: Separate into possible cases: $$\begin{align}&x+5>12\&x+5<-12\end{align}$$ - step4: Solve the inequality: $$\begin{align}&x>7\&x<-17\end{align}$$ - step5: Find the union: $$x \in \left(-\infty,-17\right)\cup \left(7,+\infty\right)$$ Solve the equation $$ |9-x| \leq \sqrt{5} $$. Solve the inequality by following steps: - step0: Solve for $$x$$: $$\left|9-x\right|\leq \sqrt{5}$$ - step1: Separate into possible cases: $$\left\{ \begin{array}{l}9-x\leq \sqrt{5}\9-x\geq -\sqrt{5}\end{array}\right.$$ - step2: Solve the inequality: $$\left\{ \begin{array}{l}x\geq -\sqrt{5}+9\x\leq \sqrt{5}+9\end{array}\right.$$ - step3: Find the intersection: $$-\sqrt{5}+9\leq x\leq \sqrt{5}+9$$ Solve the equation $$ |4+7x|=5 $$. Solve the equation by following steps: - step0: Solve for $$x$$: $$\left|4+7x\right|=5$$ - step1: Separate into possible cases: $$\begin{align}&4+7x=5\&4+7x=-5\end{align}$$ - step2: Solve the equation: $$\begin{align}&x=\frac{1}{7}\&x=-\frac{9}{7}\end{align}$$ - step3: Rewrite: $$x_{1}=-\frac{9}{7},x_{2}=\frac{1}{7}$$ Solve the equation $$ 49-7x^{2} \leq 0 $$. Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: $$49-7x^{2}\leq 0$$ - step1: Rewrite the expression: $$49-7x^{2}=0$$ - step2: Move the constant to the right side: $$-7x^{2}=0-49$$ - step3: Remove 0: $$-7x^{2}=-49$$ - step4: Change the signs: $$7x^{2}=49$$ - step5: Divide both sides: $$\frac{7x^{2}}{7}=\frac{49}{7}$$ - step6: Divide the numbers: $$x^{2}=7$$ - step7: Simplify the expression: $$x=\pm \sqrt{7}$$ - step8: Separate into possible cases: $$\begin{align}&x=\sqrt{7}\&x=-\sqrt{7}\end{align}$$ - step9: Determine the test intervals: $$\begin{align}&x<-\sqrt{7}\&-\sqrt{7}\sqrt{7}\end{align}$$ - step10: Choose a value: $$\begin{align}&x_{1}=-4\&x_{2}=0\&x_{3}=4\end{align}$$ - step11: Test the chosen value: $$\begin{align}&x<-\sqrt{7}\textrm{ }\textrm{is the solution}\&-\sqrt{7}\sqrt{7}\textrm{ }\textrm{is the solution}\end{align}$$ - step12: Include the critical value: $$\begin{align}&x\leq -\sqrt{7}\textrm{ }\textrm{is the solution}\&x\geq \sqrt{7}\textrm{ }\textrm{is the solution}\end{align}$$ - step13: The final solution is $$x \in \left(-\infty,-\sqrt{7}\right]\cup \left[\sqrt{7},+\infty\right):$$ $$x \in \left(-\infty,-\sqrt{7}\right]\cup \left[\sqrt{7},+\infty\right)$$ Solve the equation $$ |x-8|=|5x-16| $$. Solve the equation by following steps: - step0: Solve for $$x$$: $$\left|x-8\right|=\left|5x-16\right|$$ - step1: Evaluate: $$\begin{align}&x-8=5x-16\&x-8=-\left(5x-16\right)\end{align}$$ - step2: Calculate: $$\begin{align}&x=2\&x=4\end{align}$$ - step3: Rewrite: $$x_{1}=2,x_{2}=4$$ Solve the equation $$ 23-|8x|=7-x^{2} $$. Solve the equation by following steps: - step0: Solve for $$x$$: $$23-\left|8x\right|=7-x^{2}$$ - step1: Move the expression to the left side: $$23-\left|8x\right|-\left(7-x^{2}\right)=0$$ - step2: Subtract the terms: $$16-\left|8x\right|+x^{2}=0$$ - step3: Separate into possible cases: $$\begin{align}&16-8x+x^{2}=0,8x\geq 0\&16-\left(-8x\right)+x^{2}=0,8x<0\end{align}$$ - step4: Solve the equation: $$\begin{align}&x=4,x\geq 0\&x=-4,x<0\end{align}$$ - step5: Find the intersection: $$\begin{align}&x=4\&x=-4\end{align}$$ - step6: Rewrite: $$x_{1}=-4,x_{2}=4$$ Solve the equation $$ \left|x+\frac{2}{3}\right|=|-2x+3| $$. Solve the equation by following steps: - step0: Solve for $$x$$: $$\left|x+\frac{2}{3}\right|=\left|-2x+3\right|$$ - step1: Rewrite the expression: $$\left|x+\frac{2}{3}\right|=\left|2x-3\right|$$ - step2: Evaluate: $$\begin{align}&x+\frac{2}{3}=2x-3\&x+\frac{2}{3}=-\left(2x-3\right)\end{align}$$ - step3: Calculate: $$\begin{align}&x=\frac{11}{3}\&x=\frac{7}{9}\end{align}$$ - step4: Rewrite: $$x_{1}=\frac{7}{9},x_{2}=\frac{11}{3}$$ Solve the equation $$ \frac{x+9}{|x-5| x^{5}+0}<0 $$. Solve the inequality by following steps: - step0: Solve the inequality by separating into cases: $$\frac{x+9}{\left|x-5\right|\times x^{5}+0}<0$$ - step1: Find the domain: $$\frac{x+9}{\left|x-5\right|\times x^{5}+0}<0,x \in \left(-\infty,0\right)\cup \left(0,5\right)\cup \left(5,+\infty\right)$$ - step2: Simplify: $$\frac{x+9}{x^{5}\left|x-5\right|}<0$$ - step3: Separate into possible cases: $$\begin{align}&\left\{ \begin{array}{l}x+9>0\x^{5}\left|x-5\right|<0\end{array}\right.\&\left\{ \begin{array}{l}x+9<0\x^{5}\left|x-5\right|>0\end{array}\right.\end{align}$$ - step4: Solve the inequality: $$\begin{align}&\left\{ \begin{array}{l}x>-9\x<0\end{array}\right.\&\left\{ \begin{array}{l}x<-9\x \in \left(0,5\right)\cup \left(5,+\infty\right)\end{array}\right.\end{align}$$ - step5: Find the intersection: $$\begin{align}&-9\frac{3\sqrt{3}-5}{3}\end{align}$$ - step11: Choose a value: $$\begin{align}&x_{1}=-4\&x_{2}=-2\&x_{3}=1\end{align}$$ - step12: Test the chosen value: $$\begin{align}&x<-\frac{3\sqrt{3}+5}{3}\textrm{ }\textrm{is not a solution}\&-\frac{3\sqrt{3}+5}{3}\frac{3\sqrt{3}-5}{3}\textrm{ }\textrm{is not a solution}\end{align}$$ - step13: Include the critical value: $$\begin{align}&-\frac{3\sqrt{3}+5}{3}\leq x\leq \frac{3\sqrt{3}-5}{3}\textrm{ }\textrm{is the solution}\end{align}$$ - step14: The final solution is $$-\frac{3\sqrt{3}+5}{3}\leq x\leq \frac{3\sqrt{3}-5}{3}:$$ $$-\frac{3\sqrt{3}+5}{3}\leq x\leq \frac{3\sqrt{3}-5}{3}$$ ### Problema 6: Resistencia en Paralelo La ecuación que relaciona las resistencias en paralelo es: $$ \frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}} $$ Dado que $$ R_{2} = 4 $$ ohmios y $$ R $$ debe ser al menos $$ 2.5 $$ ohmios, podemos sustituir $$ R $$ en la ecuación: $$ \frac{1}{2.5} = \frac{1}{R_{1}} + \frac{1}{4} $$ Ahora, resolvemos para $$ R_{1} $$: $$ \frac{1}{R_{1}} = \frac{1}{2.5} - \frac{1}{4} $$ Calculamos $$ \frac{1}{2.5} $$ y $$ \frac{1}{4} $$: $$ \frac{1}{2.5} = 0.4 \quad \text{y} \quad \frac{1}{4} = 0.25 $$ Por lo tanto: $$ \frac{1}{R_{1}} = 0.4 - 0.25 = 0.15 $$ Finalmente, invertimos para encontrar $$ R_{1} $$: $$ R_{1} = \frac{1}{0.15} \approx 6.67 \text{ ohmios} $$ ### Problema 7: Conjuntos Solución a) Para $$ |4 + 7x| = 5 $$: $$ x_{1} = -\frac{9}{7}, \quad x_{2} = \frac{1}{7} $$ b) Para $$ |x - 8| = |5x - 16| $$: $$ x_{1} = 2, \quad x_{2} = 4 $$ c) Para $$ \left|x + \frac{2}{3}\right| = |-2x + 3| $$: $$ x_{1} = \frac{7}{9}, \quad x_{2} = \frac{11}{3} $$ d) Para $$ 23 - |8x| = 7 - x^{2} $$: $$ x_{1} = -4, \quad x_{2} = 4 $$ e) Para $$ |9 - x| \leq \sqrt{5} $$: $$ -\sqrt{5} + 9 \leq x \leq \sqrt{5} + 9 $$ f) Para $$ 12 - |x + 5| < 0 $$: $$ x \in (-\infty, -17) \cup (7, +\infty) $$ g) Para $$ 49 - 7x^{2} \leq 0 $$: $$ x \in (-\infty, -\sqrt{7}] \cup [\sqrt{7}, +\infty) $$ h) Para $$ -1 + \left(x + \frac{5}{3}\right)^{2} \leq 2 $$: $$ -\frac{3\sqrt{3} + 5}{3} \leq x \leq \frac{3\sqrt{3} - 5}{3} $$ i) Para $$ \frac{x + 9}{|x - 5| x^{5} + 0} < 0 $$: $$ -9 < x < 0 $$ ### Gráficas Para graficar los conjuntos solución, se pueden representar en la recta real los intervalos y puntos encontrados. Si necesitas las gráficas específicas, por favor indícalo. ### Problema 6: Resistencia en Paralelo Para dos resistencias $$ R_{1} $$ y $$ R_{2} = 4 $$ ohmios conectadas en paralelo, la resistencia total $$ R $$ es al menos $$ 2.5 $$ ohmios. Calculamos $$ R_{1} $$ como $$ 6.67 $$ ohmios. ### Problema 7: Conjuntos Solución a) $$ |4 + 7x| = 5 $$: $$ x = -\frac{9}{7}, \frac{1}{7} $$ b) $$ |x - 8| = |5x - 16| $$: $$ x = 2, 4 $$ c) $$ \left|x + \frac{2}{3}\right| = |-2x + 3| $$: $$ x = \frac{7}{9}, \frac{11}{3} $$ d) $$ 23 - |8x| = 7 - x^{2} $$: $$ x = -4, 4 $$ e) $$ |9 - x| \leq \sqrt{5} $$: $$ -\sqrt{5} + 9 \leq x \leq \sqrt{5} + 9 $$ f) $$ 12 - |x + 5| < 0 $$: $$ x < -17 $$ o $$ x > 7 $$ g) $$ 49 - 7x^{2} \leq 0 $$: $$ x \leq -\sqrt{7} $$ o $$ x \geq \sqrt{7} $$ h) $$ -1 + \left(x + \frac{5}{3}\right)^{2} \leq 2 $$: $$ -\frac{3\sqrt{3} + 5}{3} \leq x \leq \frac{3\sqrt{3} - 5}{3} $$ i) $$ \frac{x + 9}{|x - 5| x^{5} + 0} < 0 $$: $$ -9 < x < 0 $$ ### Gráficas Las gráficas de los conjuntos solución se representan en la recta real según los intervalos y puntos calculados.

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