Acceso
Prima
Hogar Banco de preguntas Matemáticas Álgebra
Pregunta
upstudy study bank question image url

Graph the inequality \[ 4 x+y>1 \]

Ask by Haynes Gonzalez. in the United States
Jan 26,2025

Solución de tutoría real

Respuesta verificada por el tutor

Responder

To graph the inequality \( 4x + y > 1 \), plot the line \( 4x + y = 1 \) as a dashed line. Then, shade the area above the line since it satisfies the inequality.

Solución

To graph the inequality \( 4x + y > 1 \), we can follow these steps: 1. **Convert the inequality to an equation**: First, we will consider the equation \( 4x + y = 1 \). This will help us find the boundary line for the inequality. 2. **Find the intercepts**: - **x-intercept**: Set \( y = 0 \): \[ 4x + 0 = 1 \implies x = \frac{1}{4} \] - **y-intercept**: Set \( x = 0 \): \[ 4(0) + y = 1 \implies y = 1 \] So, the intercepts are \( (0, 1) \) and \( \left(\frac{1}{4}, 0\right) \). 3. **Draw the boundary line**: Plot the points \( (0, 1) \) and \( \left(\frac{1}{4}, 0\right) \) on a graph and draw a dashed line through these points. The line is dashed because the inequality is strict (greater than, not greater than or equal to). 4. **Test a point**: To determine which side of the line to shade, we can test a point not on the line, such as \( (0, 0) \): \[ 4(0) + 0 > 1 \implies 0 > 1 \quad \text{(false)} \] Since this point does not satisfy the inequality, we shade the opposite side of the line. 5. **Final graph**: The area above the dashed line \( 4x + y = 1 \) represents the solution to the inequality \( 4x + y > 1 \). If you have graphing software or graph paper, you can plot the line and shade the appropriate region based on the steps above.

Revisado y aprobado por el equipo de tutoría de UpStudy

error msg
Explicar
Simplifique esta solución

Bonus Knowledge

To graph the inequality \( 4x + y > 1 \), start by graphing the boundary line \( 4x + y = 1 \). This can be rearranged to \( y = -4x + 1 \), a line with a slope of -4 that crosses the y-axis at (0,1). Use a dashed line for the boundary because the inequality does not include equality. Next, choose a test point not on the line, like (0,0). Plug this point into the inequality: \( 4(0) + 0 > 1 \) which simplifies to \( 0 > 1\), a false statement. This means the solution region is on the side of the line opposite to your test point (the region above the line). Shade that area to represent all the points that satisfy the inequality. Happy graphing!

preguntas relacionadas

7. Efectueaza sis scrie rezultatul sub formă de putere: \( \begin{array}{lll}\text { a) } \frac{18}{5} \cdot\left(\frac{18}{5}\right)^{2}= & \text { b) }\left(\frac{6}{5}\right)^{2} \cdot\left(\frac{6}{5}\right)^{3} \cdot \frac{6}{5}= & \text { c) }\left(\frac{19}{5}\right)^{5} \cdot\left(\frac{19}{5}\right)^{16}= \\ \begin{array}{lll}\text { d) } \frac{3}{2} \cdot\left(\frac{3}{2}\right)^{3} \cdot\left(\frac{3}{2}\right)^{0} \cdot\left(\frac{3}{2}\right)^{4}= & \text { e) }\left[\left(\frac{28}{5}\right)^{2}\right]^{3}= & \text { f) }\left[\left(\frac{5}{6}\right)^{6}\right]^{7}= \\ \text { g) }\left[\left(\frac{24}{5}\right)^{2} \cdot\left(\frac{24}{5}\right)^{3}\right]^{8}= & \text { h) }\left[\frac{5}{7} \cdot\left(\frac{5}{7}\right)^{0} \cdot\left(\frac{5}{7}\right)^{4}\right]^{5}= & \text { i) }\left(\frac{29}{10}\right)^{10}:\left(\frac{29}{10}\right)^{7}=\end{array} \\ \left.\left.\begin{array}{lll}\text { j) }\left(\frac{1}{3}\right)^{17}: \frac{1}{3}= & \left.\text { k) }\left(\frac{3}{7}\right)^{11} \cdot\left(\frac{9}{49}\right)^{3}:\left(\frac{3}{7}\right)^{15}=1\right)\end{array}\right]\left(1 \frac{1}{2}\right)^{2}\right]^{8}:\left(\frac{3}{2}\right)^{13}= \\ \text { m) }\left(\frac{9}{10}\right)^{7} \cdot\left(\frac{1}{5}\right)^{7}= & \text { n) }\left(\frac{5}{2}\right)^{10} \cdot\left(\frac{8}{5}\right)^{10}: 2^{10}= & \text { o) } 9^{3} \cdot\left(\frac{7}{10}\right)^{3}:\left(\frac{63}{10}\right)^{3}= \\ \text { p) }\left[\left(\frac{1}{5}\right)^{7}\right]^{2} \cdot 6^{14}:\left(\frac{6}{5}\right)^{14}= & \text { q) }\left(\frac{5}{2}\right)^{7}:\left(\frac{5}{2}\right)^{5}= & \end{array} \)
Álgebra Romania Jan 30, 2025
B. \( x^{2}-3 x-28 \)
Álgebra United Arab Emirates Jan 30, 2025
Oppgave 3.113 Bruk stigningstallet og konstantleddet til å tegne linjene. \( \begin{array}{ll}\text { a) } y=2 x-6 & \text { b) } y=-2 x+4\end{array} \) \( \begin{array}{ll}\text { c) } y=\frac{1}{4} x-1 & \text { d) } y=-\frac{5}{2} x+\frac{7}{2} \\ \text { av det tor to }\end{array} \)
Álgebra Norway Jan 30, 2025
Rationalize \( \frac{5}{\sqrt{2}+\sqrt{3}} \)
Álgebra Nigeria Jan 30, 2025
19. \( \frac{x^{-5} y^{4}}{y^{-3}} \)
Álgebra United States Jan 30, 2025

Latest Algebra Questions

7. Efectueaza sis scrie rezultatul sub formă de putere: \( \begin{array}{lll}\text { a) } \frac{18}{5} \cdot\left(\frac{18}{5}\right)^{2}= & \text { b) }\left(\frac{6}{5}\right)^{2} \cdot\left(\frac{6}{5}\right)^{3} \cdot \frac{6}{5}= & \text { c) }\left(\frac{19}{5}\right)^{5} \cdot\left(\frac{19}{5}\right)^{16}= \\ \begin{array}{lll}\text { d) } \frac{3}{2} \cdot\left(\frac{3}{2}\right)^{3} \cdot\left(\frac{3}{2}\right)^{0} \cdot\left(\frac{3}{2}\right)^{4}= & \text { e) }\left[\left(\frac{28}{5}\right)^{2}\right]^{3}= & \text { f) }\left[\left(\frac{5}{6}\right)^{6}\right]^{7}= \\ \text { g) }\left[\left(\frac{24}{5}\right)^{2} \cdot\left(\frac{24}{5}\right)^{3}\right]^{8}= & \text { h) }\left[\frac{5}{7} \cdot\left(\frac{5}{7}\right)^{0} \cdot\left(\frac{5}{7}\right)^{4}\right]^{5}= & \text { i) }\left(\frac{29}{10}\right)^{10}:\left(\frac{29}{10}\right)^{7}=\end{array} \\ \left.\left.\begin{array}{lll}\text { j) }\left(\frac{1}{3}\right)^{17}: \frac{1}{3}= & \left.\text { k) }\left(\frac{3}{7}\right)^{11} \cdot\left(\frac{9}{49}\right)^{3}:\left(\frac{3}{7}\right)^{15}=1\right)\end{array}\right]\left(1 \frac{1}{2}\right)^{2}\right]^{8}:\left(\frac{3}{2}\right)^{13}= \\ \text { m) }\left(\frac{9}{10}\right)^{7} \cdot\left(\frac{1}{5}\right)^{7}= & \text { n) }\left(\frac{5}{2}\right)^{10} \cdot\left(\frac{8}{5}\right)^{10}: 2^{10}= & \text { o) } 9^{3} \cdot\left(\frac{7}{10}\right)^{3}:\left(\frac{63}{10}\right)^{3}= \\ \text { p) }\left[\left(\frac{1}{5}\right)^{7}\right]^{2} \cdot 6^{14}:\left(\frac{6}{5}\right)^{14}= & \text { q) }\left(\frac{5}{2}\right)^{7}:\left(\frac{5}{2}\right)^{5}= & \end{array} \)
Álgebra Romania Jan 30, 2025
B. \( x^{2}-3 x-28 \)
Álgebra United Arab Emirates Jan 30, 2025
A. \( 3 x^{2}-x-10 \)
Álgebra United Arab Emirates Jan 30, 2025
A. \( 3 x^{2}-x-10 \)
Álgebra United Arab Emirates Jan 30, 2025
lify: \( \left(\frac{1447^{\frac{-3}{4}} 8^{\frac{2}{7}} \sqrt[6]{x^{5}} \sqrt[5]{y^{3}} \sqrt[7]{z}}{12^{12} 8^{12} 7^{2} x^{\frac{4}{5}} \sqrt[3]{y^{5}} z^{\frac{2}{6}}} * \frac{x^{\frac{-3}{4}} y^{\frac{2}{7}} z^{\frac{-8}{5}} s^{\frac{3}{2}}}{x^{2} y^{5} z^{\frac{2}{6}} s^{-9}}\right)^{\frac{7}{4}} \)
Álgebra United Arab Emirates Jan 30, 2025
Anterior Próximo
¡Prueba Premium ahora!
¡Prueba Premium y hazle a Thoth AI preguntas de matemáticas ilimitadas ahora!
Quizas mas tarde Hazte Premium
Estudiar puede ser una verdadera lucha
¿Por qué no estudiarlo en UpStudy?
Seleccione su plan a continuación
Prima

Puedes disfrutar

Empieza ahora
  • Explicaciones paso a paso
  • Tutores expertos en vivo 24/7
  • Número ilimitado de preguntas
  • Sin interrupciones
  • Acceso completo a Respuesta y Solución
  • Acceso completo al chat de PDF, al chat de UpStudy y al chat de navegación
Básico

Totalmente gratis pero limitado

  • Solución limitada
Bienvenido a ¡Estudia ahora!
Inicie sesión para continuar con el recorrido de Thoth AI Chat
Continuar con correo electrónico
O continuar con
Al hacer clic en "Iniciar sesión", acepta nuestros términos y condiciones. Términos de Uso & Política de privacidad