La somma e la differenza delle basi di un tra- pezio misurano 150 cm e 6 cm . Calcola l'area del trapezio sapendo che l'altezza è \( \frac{3}{4} \) della base minore. \( \left[4050 \mathrm{~cm}^{2}\right] \)

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.

In \( \triangle \mathrm{UVW}, u=76 \) inches, \( v=27 \) inches and \( w=83 \) inches. Find the measure of \( \angle \mathrm{U} \) to the nearest 1oth of a degree.

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

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.)

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.