Boolean/DeMorgan's Theorems

Simplifying the expression F = X̅Y + Y̅Z̅ + XY + Y̅Z yields:

• F = X + Y̅

• F = X + Y

• F = 0

• F = 1

Tags

CCSS.HSA.SSE.A.2

CCSS.HSA.SSE.B.3

______ ______

(A + B) (A̅+ B̅)

simplifies to

• A

• B

• 0

• 1

Tags

CCSS.HSA.APR.A.1

CCSS.HSA.APR.C.4

CCSS.HSA.SSE.A.2

CCSS.HSA.SSE.B.3

The Boolean expression (X + Y)(X + Y̅) simplifies to

• X

• Y

• XY

• X̅Y̅

Tags

CCSS.HSA.SSE.A.2

CCSS.HSA.SSE.B.3

Simplify the Boolean function x(x̅+y)

• xx̅

• xy

• xy̅

• x̅y̅

Tags

CCSS.HSA.SSE.A.2

CCSS.HSA.SSE.B.3

Simplify the Boolean function x + (x̅y)

• xy

• x̅ y̅

• x+y

• 0

Complete the theorem: X ⋅ X =

• 0

• 1

• X

• __

  X

Tags

CCSS.HSA.APR.A.1

Complete the theorem: X + X =

• 0

• 1

• X

• __

  X

Tags

CCSS.HSA.APR.A.1

CCSS.HSA.SSE.A.2

CCSS.HSA.SSE.B.3

Complete the theorem: X + 0 =

• X

• __

  X

• 0

• 1

Tags

CCSS.1.OA.B.3

CCSS.1.OA.D.7

__

X ⋅ X =

• 0

• 1

• X

• __

  X

Tags

CCSS.HSA.APR.A.1

A ⋅ B

is the same as

• B ⋅ A

• A ⋅ A

• B ⋅ B

• A + B

Tags

CCSS.HSA.APR.A.1