Convince your partner that segment \( B C \) is twice as long as segment \( M N \).

In \( \triangle \mathrm{STU}, u=3.6 \) inches, \( s=7.9 \) inches and \( \angle \mathrm{T}=68^{\circ} \). Find the length of \( t \), to the nearest 10th of an inch.

\( \left. \begin{array} { l } { \frac { ( 1 - h ) ^ { 2 } } { a ^ { 2 } } + \frac { ( y - k ) ^ { 2 } } { b ^ { 2 } } = 1 } \\ { \frac { h ^ { 2 } } { a ^ { 2 } } + \frac { [ ( 3 - \frac { \pi } { 2 } ) - k ] ^ { 2 } } { b ^ { 2 } } = 1 } \\ { \frac { ( - 1 - h ) ^ { 2 } } { a ^ { 2 } } + \frac { ( 1 - k ) ^ { 2 } } { b ^ { 2 } } = 1 } \\ { \frac { ( - 5 - h ) ^ { 2 } } { a ^ { 2 } } + \frac { ( 3 - k ) ^ { 2 } } { b ^ { 2 } } = 1 } \end{array} \right. \)

In \( \triangle \mathrm{EFG}, e=880 \) inches, \( f=140 \) inches and \( g=890 \) inches. Find the measure of \( \angle \mathrm{G} \) to the nearest 10th of a degree.

In \( \triangle \mathrm{TUV}, u=750 \mathrm{~cm}, v=670 \mathrm{~cm} \) and \( \angle \mathrm{T}=160^{\circ} \). Find the length of \( t \), to the nearest centimeter.

Determine the standard equation of the ellipse using the stated information. Vertices of the major axis at \( (-1,9) \) and \( (-1,-11) \); length of the minor axis is 10 units. The standard equation of the ellipse is (Simplify your answer. Use integers or fractions for any numbers in the equation.)