Let f be the function defined by f(x)=ax^2+bx+2/2x^2−8, where a and b are constants. The graph of f has a horizontal asymptote at y=3, and f has a removable discontinuity at x=2. show that a = 6. At what value of x does f have a discontinuity due to a vertical asymptote? Give a reason for your answer.

Answer: d. 39 cmin the diagram, the approximate length of the minor arc de is 39 centimetres.step-by-step explanation: the formula for the length of an arc in a circle is as displayed below.let the length of the minor arc de = xx = 2( pi )r × angle of sector / 360°from this, we will substitute the values given from the problem into the formula to obtain the length of the minor arc de.x = 2( pi )r × angle of sector / 360°r = radius = 25cmpi as an approximate value = 3.14angle of sector = 120°x = 2( 3.14 )( 25 ) × 120° / 360°x = 157 × 1 / 4x = 39.25x = 39 ( rounded to the nearest whole number as an approximate value )

Ithink it is 52

Ithink so 1.you go to your profile and your questions and you'll be able to delete them