Solve for xxx. -4x+60< 72 \quad \maroonc{\text{ or }} \quad 14x+11< -31−4x+60< 72 or 14x+11< −31minus, 4, x, plus, 60, is less than, 72, space, start color maroonc, space, o, r, space, end color maroonc, space, 14, x, plus, 11, is less than, minus, 31 choose 1 choose 1

1.The first equation is

-4x + 60<72

-4x<72-60

-4x<12

Dividing both side by -4 sign the equation reduces to

x>-3

2. The second equation is

14x+11< -31

14x+11<-31

14x<-31-11

14x<-42  

x<-3      

x<-3 or x>-3

Step-by-step explanation:

it would open into a square depending on the type of envelope i’m not sure about the second part

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)