What value(s) of b will cause 2x^2+bx+128=0 to have one real solution?

  b = -32 or +32

Step-by-step explanation:

The discriminant of the equation is ...

  d = b² -4ac

  d = b² -4(2)(128)

  d = b² -1024

In order for there to be one real solution, we must have d=0. So ...

  0 = b² -1024

  1024 = b²

  ±√1024 = ±32 = b

The value of b can be -32 or +32 to cause the equation to have one real solution.

mr. flanders’ class sold 47 doughnuts and   mr. rodriquez’s class sold 38 cartons of milk

step-by-step explanation:

let x be the number of doughnuts sold by mr. flanders’ class and y be the number of cartons sold by mr. rodriquez’s class.

1. together, the classes sold 85 items, then

x+y=85.

2. both classes earn earned $104.49 for their school. if doughnuts were sold for $1.35 each, then x doughnuts costed $1.35x. if cartons of milk were sold for $1.08 each, then y cartons costed $1.08y. hence,

1.35x+1.08y=104.49.

3. solve the system of two equations: