The diagonals of rhombus ABCD intersect at E. Given that m∠CAD=20° and CE=4, find the m∠CDA.

< CDA = 140 degrees

Step-by-step explanation:

For a rhombus, the sum of the interior angles is 360 degrees and also the opposite angles are equal i.e <ABC = <CDA and <BAD = <BCD

Given

<CAD = 20 degrees

<BCE = 20 degrees (based on geometry)

<BAD = <CAD + <BCE

<BAD = 20 + 20

<BAD = 40 degrees

Since  <BAD =  <BCD, then  <BCD = 40 degrees

Get <CDA

<CDA + <ABC + <BAD +   <BCD  = 360

<CDA + <CDA + <BAD +   <BCD  = 360

2 < CDA + 40 + 40 = 360

2< CDA + 80 = 360

2< CDA = 360-80

2< CDA = 280

< CDA = 280/2

< CDA = 140 degrees

f(x) = x^3 -2x^2 -3x

x( x^2 -2x-3)

x( x^2 +x -3x -3)

x( x(x+1) -3(x+1))

x( x-3)( x+1)

so zeroes are 0, 3,-1

its (2) option

chiken

step-by-step explanation: