bP = 1,0 = 1.48A + (0.72 * (1-A)) How does one solve for A in this given situation?
Jan 29, 2020 #1 Luclucluc 1 0 bP = 1,0 = 1.48A + (0.72 * (1-A)) How does one solve for A in this given situation?
Jan 29, 2020 #2 I like Serena Homework Helper MHB 16,350 256 Luclucluc said: bP = 1,0 = 1.48A + (0.72 * (1-A)) How does one solve for A in this given situation? What is your equation? Is it: \begin{cases} bP = 1.0 \\ 1.0 = 1.48A + (0.72 * (1-A)) \end{cases} Or is it: \begin{cases} bP = 1, \\ 0 = 1.48A + (0.72 * (1-A)) \end{cases}
Luclucluc said: bP = 1,0 = 1.48A + (0.72 * (1-A)) How does one solve for A in this given situation? What is your equation? Is it: \begin{cases} bP = 1.0 \\ 1.0 = 1.48A + (0.72 * (1-A)) \end{cases} Or is it: \begin{cases} bP = 1, \\ 0 = 1.48A + (0.72 * (1-A)) \end{cases}
Jan 29, 2020 #3 HallsofIvy Science Advisor Homework Helper 43,017 973 Assuming that your equation is [tex]1.48A+ (0.72(1- A))= 0[/tex], first expand the multiplication: [tex]0.72(1- A)= 0.72- 0.72A[/tex]. That makes the equation [tex]1.48A+ 0.72- 0.72A= 0[/tex]. Now, combine the two "A" terms: 1.48A- 0.72A= (1.48- 0.72)A= 0.76A. That makes the equation [tex]0.76A+ 0.72= 0[/tex]. Subtract 0.72 from both sides: [tex]0.76A= -0.72[/tex]. Finally, divide both sides by 0.76: [tex]A= -\frac{0.72}{0.76}= 0.9473[/tex]... Last edited by a moderator: Jan 29, 2020
Assuming that your equation is [tex]1.48A+ (0.72(1- A))= 0[/tex], first expand the multiplication: [tex]0.72(1- A)= 0.72- 0.72A[/tex]. That makes the equation [tex]1.48A+ 0.72- 0.72A= 0[/tex]. Now, combine the two "A" terms: 1.48A- 0.72A= (1.48- 0.72)A= 0.76A. That makes the equation [tex]0.76A+ 0.72= 0[/tex]. Subtract 0.72 from both sides: [tex]0.76A= -0.72[/tex]. Finally, divide both sides by 0.76: [tex]A= -\frac{0.72}{0.76}= 0.9473[/tex]...