43÷1454 help me solve please ​

3 is greater then both of those numbers

i'm sorry mate, but i am not seeing any figures. i am unable to solve this for you.

< 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)

this is a direct variation proportion. y = kx

y is the total amount due, k is the constant of variation, and x is the number of texts sent.

plug in the 3.20 and the 16 to solve for the "k". this is the constant of proportionality. it is the value that will not change.

3.20 = k(16). solving you get $.20 per text.

the equation would then be y = .20x