Find the \( z \) value to the left of the mean so that \( 90.82 \% \) of the area under the distribution curve lies to the right of it. Use The Standard Normal Distribution Table and enter the answer to 2 decimal places. The

\( H(x)=\frac{x^{3}-64}{x^{2}-5 x+4} \) Find the vertical asymptotes. Select the correct choice and, if necessary, fill in the answer box(es) to complete your choice. A. The function has one vertical asymptote, (Type an equation. Use integers or fractions for any numbers in the equation.) B. The function has two vertical asymptotes. The leftmost asymptote is (Type equations. Use integers or fractions for any numbers in the equations.) C. The function has no vertical asymptote.

Solve the equation. Identify excluded values as appropriate. \[ -\frac{5}{3} h=9 \]

Si \( g(x)=\frac{3 x^{2}+x+1}{4 x-x^{2}} \) entonces \( g(-x)= \)

a) Determine whether each of these statements is true or false. i. \( \quad \emptyset \in\{0\} \) False ii. \( \{0\} \subset \emptyset \) False iii. \( \emptyset \subset\{0\} \) True iv. \( x \in\{x\} \) True v. \( \{x\} \subseteq\{x\} \) True vi. \( \{x\} \in\{x\} \) False vii. \( \{x\} \in\{\{x\}\} \) True viii. \( \emptyset \subseteq\{x\} \) True

\( \left. \begin{array} { c } { ( x _ { 1 } , y _ { 1 } , z _ { 1 } ) \oplus ( x _ { 2 } , y _ { 2 } , z _ { 2 } ) = ( x _ { 1 } + x _ { 2 } , y _ { 1 } + y _ { 2 } , z _ { 1 } + z _ { 2 } ) } \\ { \alpha \odot ( x , y , z ) = ( \alpha x , \alpha y , \alpha z ) } \end{array} \right. \)