Fifteen years ago phil was twice as old as darell was then. three years from now darell will be as old as phil was four years ago. how old is phil now

29years

Step-by-step explanation:

let phil age now = x

15 years ago, phil age = x-15

let darell age now = y

15 years ago darell age = y-15

15 years ago,

phil's age = twice darell's age

x-15 = 2(y-15)

x-15 =2y-30

x-2y = -30+15

x-2y = -15 equation (1)

four years ago, Phil's age = x-4

three years from now, darell's age = y+3

x-4=y+3

x-y =3+4

x-y =7

x=7+yequation(2)

subtituting the value of x in equation (2) to equation (1)

x-2y = -15

7+y -2y = -15

7-y=-15

y=7+15=22

subtituting the value of y into equation (2)

x=7+y

=7+22=29

therefore, Phil age now is 29years

b: |12 – x| ≤ 0.25

step-by-step explanation:      

let x be the possible lengths in inches, of the ribbons jaime cut.

we have been given that jaime is cutting strips of ribbon for a project. he tries to cut each piece to be 12 inches long, but sometimes he cuts it too long or too short by as much as 0.25 inch.  

we can represent this information in an inequality as:

by the definition of absolute value function we will get,

upon combining our inequalities we will get,

therefore, the inequality   describes all the possible lengths (x), in inches, of the ribbons jaime cut.

step-by-step explanation:

step-by-step explanation:

we can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it. (note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 v, i.e. w = 1 ev = 1.6 10-19 c 1 v = 1.6 10-19j).in fact this expression is really nothing more than the statement that potential is the potential energy per unit charge. in this case, v= w/qis simply stating that the change in potential energy(or work done) per unit charge is equal to the change in potential (or potential difference), v.student questions: uniform electric fields2electron beamcathode+5 kvanode0 v