Mike ran 2 km in 13 minutes.  If he continues at this same pace, how long will it take Mike to run 7 km?  Round your answer to t

Answer:

Part 1) The equation is

Part 2) When x=40 m, the area of the schoolyard is A=2,800 m^2

Part 3) The valid domain consists of all real numbers exceeding zero and below 75 meters

Step-by-step explanation:

Part 1) Formulate an expression for A(x)

Let

x -----> the length of the rectangular school yard

y ---> the width of the rectangular school yard

It is known that

The perimeter for the fencing (taking the school wall as one side) is

thus

-----> this is equation A

The area of the rectangular school yard is

----> this is equation B

Substituting equation A into equation B yields

Change to function notation

Part 2) What is the area when x=40?

With x equal to 40 m

substitute the value into the expression from Part 1 to determine A

   

Part 3) What would be a suitable domain for A(x) in this scenario?

We understand that

A signifies the area of the rectangular school yard

x characterizes the length of the rectangular school yard

It follows that

This forms a vertical parabola opening downwards

The vertex indicates a maximum point

The x-coordinate of the vertex corresponds to the length that maximizes the area

The y-coordinate of the vertex denotes the maximum area

The vertex corresponds to (37.5, 2812.5)

Refer to the accompanying figure

Consequently,

The peak area achieved is 2,812.5 m^2

The x-intercepts are located at x=0 m and x=75 m

The domain for A is the range -----> (0, 75)

All real numbers greater than zero and less than 75 meters

Answer:

2.55 cubic meters of sand

Step-by-step explanation:

The inquiry focuses on how much sand is needed to completely fill the sandbox. As it is a three-dimensional shape, the concept of 'Volume' is applicable. Volume is determined through the formula L × W × H. By substituting the relevant variables into this equation, you arrive at the final solution. I hope this offers clarity!

Each LED bulb, along with installation labor, is priced at
.. $6.95 +$3 = $9.95

For 100 bulbs over a span of 10 years, that equals (100*10) = 1000 bulb·years. At $9.95 per bulb, 5 bulb·years are obtained, and thus the projected total cost for 1000 bulb·years is
.. (1000 b·y)*($9.95/(5 b·y)) = $1990

In summary, for a decade, the installation and changes of 200 bulbs in 100 lamps amount to $1990. Therefore, the yearly cost is...
.. $1990/(10 yr) = $199/yr

The dimensions are 58 ft × 58 ft. Step-by-step explanation: Let the length of the region be represented as x feet, and the width as y feet. Given a perimeter of 234 feet, the area A can be represented as xy. By differentiating the equation with regards to x, we can determine the point of maximum area, revealing that for x = 58.5 feet, the area's maximum occurs when both dimensions are 58.5 ft.