It takes \( 8 \) minutes for Byron to fill the kiddie pool in the backyard using only a handheld hose. When his younger sister is impatient, Byron also uses the lawn sprinkler to add water to the pool so it is filled more quickly. If the hose and sprinkler are used together, it takes \( 5 \) minutes to fill the pool. Which equation can be used to determine r; the rate in parts per minute, at which the lawn sprinkler would fill the pool if used alone? 

\( \frac { 5 } { 8 } + 5 r = 8 \) 

\( \frac { 5 } { 8 } + 5 r = 1 \) 

\(5 ( \frac { 5 } { 8 } ) = r \) 

\( \frac { 5 } { 8 } = 5 r\) 

Find such that the function \(f( x) = \left \{ \begin{array} { rcl} 2x^ 2 + 4x & \text { if } & x \geq 1 \\  - x + k & \text { if } & x < 1 \end{array} \right . \)is continuous on \(\mathbb { R} \).

Market research for a certain ice cream company indicates that if the price per cup of ice cream is \(P 50.00\) , the demand will be \(1200\) cups, whereas if the price is increased to \(P 60.00\) , the demand will be \(1100\) cups. 

(a) If \(p\) denotes the price per cup (in pesos) and \(x\) denotes the demand for ice cream (in cups), express \(p\) as a linear function of \(x\) . 

(b) Determine the revenue function and its domain. 

(c) Find the marginal revenue at a production level of \(200\) cups of ice cream. Interpret. 

Now, suppose it is known that the total cost of production of \(x\) cups of ice cream is given by ( \(x ) = 3 x + 25,500\) . 

(d) Find the value(s) of \(x\) at the break-even points. 

(e) Use marginal analysis to approximate the profit from the sale of the \(500 ^ { th} \) cup of ice cream. Compare this value with the exact profit from the sale of the \(500 ^ { th} \) cup. 

Write the system of equations corresponding to the given augmented matrix using the variables \(x\) and \(y\) . Then perform the row operation \(R _ { 2 } = - 7 r _ { 1 } + r _ { 2 } \) on the given augmented matrix. 

\(\begin{bmatrix}    1 & 2 &\big m| & 9 \\    7 & 8 &\big m| & - 1 \\    \end{bmatrix} \) 

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 added to the account 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 became a competitive disadvantage for financial institutions, and some began to divide the annual interest into periodic shares, so that (for example) you could get \(1/12^ { th} \) of that \(6 \% \) each month. When this happens, we say that interest is compounded monthly. Interest can also be compounded weekly (52 times per year), quarterly ( \(4\) times per year), daily ( \(365\) times per year), or really any other period you could think of. 

If interest is compounded monthly, what growth factor would be needed to provide \(1/12^ { th} \) of \(6 \% \) interest each month? (Think about the difference between interest rate and growth factor.) 

The growth factor would be \(\square \) 

1. Use the definition of the definite integral to evaluate the integral. Use the right end point of each interval for \(x_ { \space i} ^ *\).

\[\int _ { 1} ^ { 4} { { 2x + 3\space dx} } \]