If |vec{f}|=10, |vec{g}|=14 and |vec{f}-vec{g | vedclass

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(B) We are given $|\vec{f}|=10$,$|\vec{g}|=14$,and $|\vec{f}-\vec{g}|=15$. Using the property $|\vec{f}-\vec{g}|^2 = |\vec{f}|^2 + |\vec{g}|^2 - 2(\ve

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$\sqrt{367}$

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(B) We are given $|\vec{f}|=10$,$|\vec{g}|=14$,and $|\vec{f}-\vec{g}|=15$. Using the property $|\vec{f}-\vec{g}|^2 = |\vec{f}|^2 + |\vec{g}|^2 - 2(\vec{f} \cdot \vec{g})$,we have: $15^2 = 10^2 + 14^2 - 2(\vec{f} \cdot \vec{g})$ $225 = 100 + 196 - 2(\vec{f} \cdot \vec{g})$ $225 = 296 - 2(\vec{f} \cdot \vec{g})$ $2(\vec{f} \cdot \vec{g}) = 296 - 225 = 71$. Now,we need to find $|\vec{f}+\vec{g}|$. Using the property $|\vec{f}+\vec{g}|^2 = |\vec{f}|^2 + |\vec{g}|^2 + 2(\vec{f} \cdot \vec{g})$,we substitute the known values: $|\vec{f}+\vec{g}|^2 = 10^2 + 14^2 + 71$ $|\vec{f}+\vec{g}|^2 = 100 + 196 + 71 = 367$ Therefore,$|\vec{f}+\vec{g}| = \sqrt{367}$.

If $|\vec{f}|=10, |\vec{g}|=14$ and $|\vec{f}-\vec{g}|=15$,then $|\vec{f}+\vec{g}|=$

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