2.5 Write an equation using a letter symbol for the unknown number to describe the problem in each situation below. 2.5.1 There are 43 passengers on a bus. At a bus stop \( m \) passengers get on. There are now 70 passengers on the bus. 2.5.2 Moses is 26 years old. His father is 3 years less than three times Moses's age.

Trovare i valori dei parametri reali \( a, b \in \mathbb{R} \) che rendono continua la seguente funzione \[ f(x)=\left\{\begin{array}{ll}\frac{2 e^{1-\cos (\sqrt{x})}+\sin (x)-2}{x} & \text { se } x>0 \\ b x e^{a x}+a & \text { se } x \leq 0\end{array}\right. \]

\( \left. \begin{array} { | l | l | l | l | l | l | l | } \hline 5 & { 174 } & { 234 } & { 456 } & { 275 } & { 193 } \\ { 2 } & { \times 3 } & { \times } & { 4 } & { \times 2 } & { \times 3 } & { \times 4 } \\ { 0 } & { 2 } & { 3 } \end{array} \right. \)

-4. \( 3+4 m=-7 \)

An equation of the form \( y=6000(1.06)^{x} \) provides an example of interest compounded annually. This means that the full \( 6 \% \) of interest is a at the end of one year. This doesn't sound very fair to someone that invests their money for 11 months-they get no interest at all. This beca disadvantage for financial institutions, and some began to divide the annual interest into periodic shares, so that (for example) you could gel each month. When this happens, we say that interest is compounded monthly. Interest can also be compounded weekly ( 52 times per year), per year), daily ( 365 times per year), or really any other period you could think of. If interest is compounded monthly, calculate how many times the growth factor would be applied in each of the following time periods. Part: 0 / 3 Part 1 of 3 (a) If interest is compounded monthly, how many times would the growth factor be applied in 2 years? The growth factor will need to be applied

Find \( f^{\prime}(x) \) \[ f(x)=8 e^{-2 x} \]