Find the value of x^3 + y^3 + 12xy – 64, when x +y = 4

0

Step-by-step explanation:

Using the algebraic identity

(x + y)³ = x³ + y³ + 3xy(x + y) ← substitute x + y = 4

4³ = x³ + y³ + 12xy , that is

x³ + y³ + 12xy = 64 ( subtract 64 from both sides )

x³ + y³ + 12xy - 64 = 0

the answer is zero

Step-by-step explanation:

x^3 + y^3 + 12xy – 64 factorize the polynomial

x^3+y^3-4^3+3(4xy)      64 is 4^3

take x+y-4 as a common factor

(x+y-4)(x^2-xy+4x+y^2+4y+16) substitute x+y=4

(4-4)(x^2+y^2-4^2-3xy)

=0 any amount multiply by zero =0

i can you, i love poetry

a) one student is chosen from classroom a, then a student is chosen from classroom b.  

step-by-step explanation:

independent events are ones in which the probability of one event does not affect the probability of the second event.

one student is chosen from classroom a; this does not affect the chances of choosing a student from classroom b.   this makes these independent events.

when one card is chosen from a standard deck and then set aside before a second card is chosen, the probability of the second card being chosen changes.   this makes them dependent events.

when one student is chosen from classroom a and then a second student is chosen from classroom a, the probability of choosing the second student changes.   this makes them dependent events.

when a letter is chosen from the alphabet and then a second letter is chosen, the probability of choosing the second letter changes.   this makes them dependent events.