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11Alexandr11 [23.1K]
3 years ago
11

At 9:00 a.m. marsha found a parking meter that still had 5 minute until it expired. She quickly put a quarter, 1 dime, and a nic

kel into The meter and went to her meeting. If 5¢ buys 15 minute of parking time, at What time will The meter expire
Mathematics
1 answer:
LUCKY_DIMON [66]3 years ago
8 0
1 quarter + 1 dime + 1 nickel = 25¢ + 10¢ + 5¢ = 40¢

Each 5<span>¢ buys 15 minutes.  How many ' 5</span>¢ ' are in 40<span>¢</span> ?

40/5 = 8 of them.  So the (quarter+dime+nickel) can buy 8 x 15 minutes = 2 hours.

Starting at 9:00 AM, the meter had 5 minutes left on it, and Marsha bought
another 2 hours.  So now the meter is good until 2 hours and 5 minutes
after 9:00 AM . . . . . 11:05 AM.






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We know that :

⊕  Sum of the interior angles in a Pentagon should be equal to 540°

⇒  x° + (2x)° + (2x)° + 90° + 90° = 540°

⇒  (5x)° = 540° - 180°

⇒  (5x)° = 360°

\sf{\implies x^{\circ} = \dfrac{360^{\circ}}{5}}

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Questions in Image below. Please explain how you got your answers in depth please.
irakobra [83]

Step 1

Revenue (sometimes referred to as sales revenue) is the amount of gross income produced through sales of products or services. This, in the context of the question, means that the revenue generated from the sales of tickets is given by the function below. We are now required to find the price that tickets will be sold at in a day based on the function that will make the revenue to be equal to $0.

The revenue is given by the function;

\begin{gathered} f(p)=150p-5p^2 \\ f(p)\text{ is the revenue and should be equal to 0 according to the question.} \\ p=\text{ The price of tickets for the day for the revenue to be 0} \end{gathered}

To solve this problem or to find this price at which tickets will be solved to give us zero revenue, we must equate the given function for the revenue to 0 and find the value of p, the price of the tickets that make the function to be equal to 0.

For the theatre to make $0 in revenue, this means f(p)=0.

Therefore, we will equate f(p) to 0

Step 2

Equate f(p) to 0

\begin{gathered} 0=150p-5p^2 \\ p(150-5p)=0----(\text{factorize)} \\ p=0 \\ 0r \\ 150-5p=0 \\ 150=5p \\ \frac{5p}{5}=\frac{150}{5} \\ p=30 \\ p=0\text{ or 30} \end{gathered}

Therefore the revenue to be zero, the price of the ticket will be $0 or $30 but the question asked us how high the ticket price will be to get a revenue of 0. Both 0 and 30 gave us revenue of 0 but $30 is higher therefore, the answer will be $30.

Check;

\begin{gathered} 150p-p^2=0 \\ \text{If the price of tickets=0} \\ 150(0)-5(0^2)=0 \\ \text{If p=30} \\ 150(30)-5(30^2)=4500-4500=0 \end{gathered}

Both give us 0 revenue but $30 is higher and the right answer.

Step 3

The equation you can write for revenue of $700 is;

\begin{gathered} 700=150p-5p^2 \\ 5p^2-150p+700=0_{} \end{gathered}

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