If $10,000 is placed in an investment paying 4.8% yearly interest compounded annually, how long will it take for the account to double?

15 years.

Step-by-step explanation:

We have the equation:

20000 = 10000(1 + 0.048)^t     where t is the number of years.

1.048 t = 20000/10000 = 2

Taking logs:

t ln 1.048 = ln 2

t = ln 2 / ln 1.048

t = 14.78

< 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)