Practicing Success
Practicing Success
CUET Preparation Today
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a b c d |
c |
so $C=-ln|50|$ and $-ln|50-y|=kx-ln|50|$ $ln|50-y|-ln|50|=-kx$ so $ln\left|1-\frac{y}{50}\right|=-kx$ $1-\frac{y}{50}=e^{-kx}$ so $y=50(1-e^{-kx})$ |
