Answer:
Step-by-step explanation:
![y^3-6y^2+5y=y(y^2-6y+5)=y(y^2-y-5y+5)\\\\=y[y(y-1)-5(y-1)]=y(y-1)(y-5)](https://tex.z-dn.net/?f=y%5E3-6y%5E2%2B5y%3Dy%28y%5E2-6y%2B5%29%3Dy%28y%5E2-y-5y%2B5%29%5C%5C%5C%5C%3Dy%5By%28y-1%29-5%28y-1%29%5D%3Dy%28y-1%29%28y-5%29)
The tool is easy to use. All you need to do is choose the number of words you want to create (the default is five, but you can input any number you'd like) and the type of words you want. You can choose from all words, verbs only, nouns only or adjective only depending on which best meets your needs.
Sorry im only in middle school i dont know this
Answer:
but dude what to answer............
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!!!
