Answer:
b = -2c ± [√(4π²c² + πA)]/π
Step-by-step explanation:
A = 4πbc + πb^2
A = 4πbc + πb²
πb² + 4πbc - A = 0
Using the quadratic formula to solve this quadratic equation.
The quadratic formula for the quadratic equation, pb² + qb + r = 0, is given as
b = [-q ± √(q² - 4pr)] ÷ 2p
Comparing
πb² + 4πbc - A = 0 with pb² + qb + r = 0,
p = π
q = 4πc
r = -A
b = [-q ± √(q² - 4pr)] ÷ 2p
b = {-4πc ± √[(4πc)² - 4(π)(-A)]} ÷ 2π
b = {-4πc ± √[16π²c² + 4πA]} ÷ 2π
b = (-4πc/2π) ± {√[16π²c² + 4πA] ÷ 2π}
b = -2c ± [√(4π²c² + πA)]/π
Hope this Helps!!!
Y=-4x-1. Since the 1 has to be negative since the equation is y=mx-b. M is the slope and b is the y intercept
Answer: u = -4w + 9
<u>Step-by-step explanation:</u>
u - 5 = -4(w - 1)
u - 5 = -4w + 4
<u> +5 </u> <u> +5 </u>
u = -4w + 9
