What is the sum of 23/25 + 9/30

61/50

Step-by-step explanation:

1*a = a

1 = b/b

then:

23*25 * 1 = (23/25)(12/12) = (23*12)/(25*12) =276/300

9/30 *1 = (9/30)(10/10) = (9*10)/(30*10) = 90/300

then:

(23/25) + (9/30) = (276/300) + (90/300)

= (276+90)/300

= 366/300

= 122/100

= 61/50

< 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)