The curvature of the helix r(t)equals(a cosine t )iplus(a sine t )jplusbt k (a, bgreater than or equals0) is kappaequalsstartfraction a over a squared plus b squared endfraction . what is the largest value kappa can have for a given value of b?

i don't even know man. edge inuity be messing with us. i had to ask a question too.

x = 17

step-by-step explanation:

1st step: isolate x.

24 - 3x = -27

-24         -24

-3x   =   -51

2nd step: divide.

-3x     =   -51

  3             3

  x       =     17

hope this

correct answer is c. just did it

step-by-step explanation:

x = 2

step-by-step explanation:

if sr parallel to ba then the sides bc and ac are in proportion, that is

= ← substitute values

= ( cross- multiply )

27x = 6(4x + 1)

27x = 24x + 6 ( subtract 24x from both sides )

3x = 6 ( divide both sides by 3 )

x = 2