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2 years ago

If it costs $100,000 to put on an event for four weeks (28 consecutive nights), how much revenue per night is needed to

Mathematics

1 answer:

Arada [10]2 years ago

8 0

100000+200000=300000
28x=300000
X=10714
They would need to make $10714 a night

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Answer:

1/3

Step-by-step explanation:

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2 years ago

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Use completing the square to solve this quadratic equation x^2+10x+25=2

never [62]

answer:

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2 years ago

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Julli [10]

Answer:

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3 years ago

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Dylan drank 5.2 ounces of water before soccer practice, and 12.9 ounces after practice. Before he drank any water, there were 28

Elan Coil [88]

Answer:

$x=28-5.2-12.9$ , 9.9 ounces of water remaining.

Step-by-step explanation:

Given:

Dylan drank 5.2 ounces of water before soccer practice, and 12.9 ounces after practice.

Before he drank any water, there were 28 ounces of water in the container.

Question asked:

Which expression can be used to find the amount of water remaining?

Solution:

Let $x$ the amount of water remaining.

Total quantity of water in the container = 28 ounces

Before practice, Dylan drank = 5.2 ounces

After practice, Dylan drank = 12.9 ounces

Remaining amount of water = Total quantity of water in the container - Amount of water is drunk before practice - Amount of water is drunk before practice

$x=28-5.2-12.9\\ \\ x=28-18.1\\ \\ x=9.9\ ounces$

Thus, 9.9 ounces of water remaining in the container.

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3 years ago

Let’s pretend that we want to give our employees in total a28% raise in the next two years, but we want to spread out aconsisten

Roman55 [17]

Solution

- The solution steps are given below:

$\begin{gathered} \text{ Let the original amount be X} \\ \\ \text{ If we want a 28\% raise over the next two years, then it means that the amount will be:} \\ X+\frac{28}{100}X=1.28X \\ \\ \text{ If we want to give them a consistent raise each year, we can compute the scenario as follows:} \\ \text{ Let the percentage be }y \\ \\ X+y\%\text{ of }X=\text{ Salary after first year} \\ \\ (X+y\%\text{ of }X)+y\%\text{ of }(X+y\%\text{ of }X)\text{ = Salary after second year.} \\ \\ \text{ But we already know that the salary after second year is }1.28X \\ \text{ Thus, we can say:} \\ (X+y\%\text{ of }X)+y\%\text{ of }(X+y\%\text{ of }X)=1.28X \\ \\ \text{ Simplifying, we have:} \\ X+\frac{yX}{100}+\frac{y}{100}(X+\frac{yX}{100})=1.28X \\ \\ \text{ Divide through by }X \\ 1+\frac{y}{100}+\frac{y}{100}(1+\frac{y}{100})=1.28 \\ \\ \text{ Subtract 1 from both sides and expand the brackets} \\ \frac{y}{100}+\frac{y}{100}+(\frac{y}{100})^2=1.28-1=0.28 \\ \\ \frac{2y}{100}+(\frac{y}{100})^2=0.28 \\ \\ \text{ Multiply both sides by }100^2 \\ 200y+y^2=2800 \\ \\ \text{ Rewrite, we have:} \\ y^2+200y-2800=0 \\ \\ Solving\text{ the equation using the Quadratic formula, we have that:} \\ y=\frac{-200\pm\sqrt{200^2-4(-2800)}(1)}{2(1)} \\ \\ y=-213.137\text{ or }13.137 \\ \\ \text{ Since the change in salary is an increase, thus, the rate }y\text{ has to be positive.} \\ \text{ Thus, } \\ y=13.137\approx13.14\% \\ \\ \end{gathered}$

Final Answer

y = 13.14%

The screenshots of the solution are:

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6 months ago