- {-b+{\
sqrt {b^{2}-4ac}}}{2a}},\qquad x_{2}={\frac {-b-{\
sqrt {b^{2}-4ac}}}{2a}}.} The
quantity Δ = b 2 − 4 a c {\displaystyle \textstyle \
Delta =b^{2}-4ac}...
- approximation. The
delta method therefore implies that n ( h ( B ) − h ( β ) ) → D N ( 0 , ∇ h ( β ) T ⋅ Σ ⋅ ∇ h ( β ) ) {\displaystyle {\
sqrt {n}}\left(h(B)-h(\beta...
- {\displaystyle {\begin{cases}{\
sqrt {2\
Delta R^{2}+4\
Delta G^{2}+3\
Delta B^{2}}},&{\bar {R}}<128,\\{\
sqrt {3\
Delta R^{2}+4\
Delta G^{2}+2\
Delta B^{2}}}&{\text{otherwise}}...
- _{1});\\\
Delta {Z}&=\sin(\phi _{2})-\sin(\phi _{1});\\D_{\textrm {t}}&=R{\
sqrt {(\
Delta {X})^{2}+(\
Delta {Y})^{2}+(\
Delta {Z})^{2}}}\\&=2R{\
sqrt {\sin ^{2}{\frac...
- b − Δ 2 a , {\displaystyle {\frac {-b+{\
sqrt {\
Delta }}}{2a}}\quad {\text{and}}\quad {\frac {-b-{\
sqrt {\
Delta }}}{2a}},} both of
which are real numbers...
- {1}{2}}{\
sqrt {-{\frac {2}{3}}\ p+{\frac {1}{3a}}\left(Q+{\frac {\
Delta _{0}}{Q}}\right)}}\\Q&={\
sqrt[{3}]{\frac {\
Delta _{1}+{\
sqrt {\
Delta _{1}^{2}-4\
Delta...
- ^{0}={\frac {\
sqrt {\varepsilon \
Delta }}{\rho }}\left(dt-a\sin ^{2}\theta \,d\phi \right)} σ 1 = ρ ε Δ d r {\displaystyle \sigma ^{1}={\frac {\rho }{\
sqrt {\varepsilon...
- {\displaystyle C={\
sqrt[{3}]{\frac {\
Delta _{1}\pm {\
sqrt {\
Delta _{1}^{2}-4\
Delta _{0}^{3}}}}{2}}},}
where the
symbols {\displaystyle {\
sqrt {{~}^{~}}}}...
- + Δ z 2 − c 2 Δ t 2 , {\displaystyle \
Delta \sigma ={\
sqrt {\
Delta x^{2}+\
Delta y^{2}+\
Delta z^{2}-c^{2}\
Delta t^{2}}},}
where Δt is the
difference in...
-
Delta _{1}+\
Delta _{2}\right)\left(\left[\
Delta _{2}E_{f}+\
Delta _{1}p_{f}c\right]-{\
sqrt {\left[\
Delta _{2}E_{f}+\
Delta...
