If xy + 2ey = 2e, find the value of y'' at the point where x = 0.

y = 1 is the value at x = 0

Step-by-step explanation:

The given equation is xy + 2ey = 2e

We have to solve this equation to find the value of y at the point x = 0.

xy + 2ey = 2e

y(x + 2e) = 2e

Now we divide this equation by (x + 2e) on both the sides of the equation.

Now we have find the value of y at x = 0

By putting x = 0 in the simplified form of the equation

So the value of y is 1.

step-by-step explanation:

the multiplicative inverse of a fraction a/b is b/a. for the multiplicative inverse of a real number, divide 1 by the number. for example, the reciprocal of 5 is one fifth 1/5 or 0.2, and the reciprocal of 0.25 is 1 divided by 0.25, or 4.

< 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)