c) \( \frac{3^{7} \cdot 10^{8}}{3^{2} \cdot 10^{4}}= \)

b. \( \frac{1}{2}+\frac{\square}{4}-\frac{3}{2}=\frac{5}{4} \)

(7) Rodea las fracciones equivalentes entre si a. \( 1 \frac{1}{2} \quad \frac{5}{4} \quad 1+\frac{1}{4}+\frac{1}{4} \quad \frac{3}{8}+1 \)

\( \left[\begin{array}{ll}x & 7 y \\ z & 4\end{array}\right]=\left[\begin{array}{cc}16 & 28 \\ 13 & 4\end{array}\right] \) \( x=16 \) (Simplify your answer.) \( y=\square \) (Simplify your answer.)

(1 point) Monday's lecture went over a logistic equation: \( f(t)=\frac{1000}{1+999 e^{-0.603 t}} \) which models how many people \( f \) have had the flu \( t \) days after an initial infection in a community of 1000 people. According to the model, how many people will have had the flu after 10 days? 294 According to the model, how many people will have had the flu after 20 days? According to the model, how many days until 500 people have had the flu? 11.45 According to the model, how many days until 700 people have had the flu? According to the model, how many days until 2000 people have had the flu? never Hint: One of your answers should be "never"

Dxercice 2: (1) Montrer les implications suivantes: a \( (\forall x, y \in \mathbb{R}) ; \quad \mathrm{Si} \quad(|\mathrm{x}| \leq 1 \) et \( |y| \leq 1) \quad \) alors \( \sqrt{1-\mathrm{x}^{2}}+\sqrt{1-\mathrm{y}^{2}} \leq 2 \sqrt{1-\left(\frac{\mathrm{x}+\mathrm{y}}{2}\right)^{2}} \) b \( \left(\forall \mathrm{x}, \mathrm{y} \in \mathbb{R}^{*}\right) ; \quad 2 \mathrm{x}+4 \mathrm{y}=1 \Rightarrow \frac{1}{\mathrm{x}^{2}+\mathrm{y}^{2}} \leq 20 \) (2) En utilisant "Le raisonnement par les équivalences successives", montrer que: a \( \left(\forall \mathrm{x}, \mathrm{y} \in \mathbb{R}^{+}\right) ; \quad \mathrm{x} \sqrt{\mathrm{x}^{2}+1}=y \sqrt{\mathrm{y}^{2}+1} \Longleftrightarrow x=y \) \( \mathrm{~b}\left(\forall \mathrm{x}, \mathrm{y} \in \mathbb{R}_{+}^{*}\right) ; \quad \mathrm{x} \prec \mathrm{y} \Longleftrightarrow \frac{\mathrm{x}}{\mathrm{y}} \prec \frac{2 \mathrm{x}+5 \mathrm{y}}{5 \mathrm{x}+2 \mathrm{y}} \prec \frac{\mathrm{y}}{\mathrm{x}} \)