Hello! I am trying to finish a math assignment. it was a bunch of math questions, and I am having a very difficult time with 5 of them. I have the answers to them though. Could somebody please help me with AT LEAST a few!? thanks!!

1) A square swimming pool with a side measuring 16m is to be surrounded by a uniform rubberized floor covering. if the area of the floor covering equals the area of the pool, find the width of the rubberized covering. (answer 3.3m)

2) A rectangular area is enclosed by a fence and divided by another fence parallel to two of its sides. if the 600m of fence used encloses a maximum area, what are the dimmensions of the enclosure? ( answer 100 by 150)

3) A theatre seats 2000 people and charges 10$ for a ticket. at this price, all tickets will be sold. a survey indicates that if the ticket price is increased, the number sold will decrease by 100 for every dollar of increase. what ticket price will result in the greatest revenue? (answer 15$)

4) To fave fuel on the 240km trip to the cottage, the Nakamura family reduced their usual average speed by 20 km/h. this lengthens their journey by 1 hour. What is the slower average speed? ( answer 60 km/h)

5) the edges of 3 cubes are consecutive odd integers. if the cubes are stacked on a desk as shown, the total exposed surface area is 281 cm squared. find the lenghts of the sides of the cubes. ( answer 3 cm, 5 cm, 7cm)

i have tried countless diagrams and charts and i feel really dumb cuz i can't get anything. help would be much appreciated! thanks!

Of course, I'd be happy to help you with these math questions! Let's go through each question step by step:

1) For the first question, we have a square swimming pool with a side length of 16m. We want to find the width of the rubberized covering that surrounds the pool. Since the area of the covering equals the area of the pool, we can set up the equation:

Area of covering = Area of pool

Let's denote the width of the rubberized covering as 'x'. The area of the pool is equal to the side length squared, which is 16^2 = 256m^2. The area of the covering is (16 + 2x)^2 (since the covering goes around all sides of the pool, we add twice the width to each side).

Now we can set up the equation:

256m^2 = (16 + 2x)^2

To solve for x, we need to take the square root of both sides and then solve the resulting quadratic equation. After simplifying, we find that x ≈ 3.3m.

2) For the second question, we have a rectangular area that is enclosed by a fence and divided by another fence parallel to two of its sides. We want to find the dimensions of the enclosure that maximize the area. Let's denote the width of the enclosure as 'x' and the length as 'y'.

We know that the perimeter of the enclosure is 600m. Since we have two equal sides (due to the parallel fence), we can set up the equation:

2x + 3y = 600

To maximize the area, we need to express the area in terms of one variable. The area of a rectangle is given by A = x * y. Now, we can express y in terms of x using the equation for perimeter:

y = (600 - 2x) / 3

Substituting this into the area equation, we get:

A = x * (600 - 2x) / 3

To find the maximum area, we can take the derivative of A with respect to x, set it equal to zero, and solve for x. After solving, we find that x = 100m and y = 150m.

3) For the third question, we want to find the ticket price that will result in the greatest revenue for the theater. We know that at $10 per ticket, all 2000 tickets will be sold. If the ticket price is increased, the number of tickets sold will decrease by 100 for every dollar increase in price.

Let's denote the ticket price as 'x' in dollars and the number of tickets sold as 'y'. We want to maximize the revenue, which is given by R = x * y.

We know that when the price is $10, y = 2000. For each dollar increase in price, the number of tickets sold decreases by 100. So the equation for the number of tickets sold, y, as a function of the ticket price, x, is:

y = 2000 - 100(x - 10)

To find the revenue function, we multiply the ticket price by the number of tickets sold:

R = x * (2000 - 100(x - 10))

To find the ticket price that maximizes revenue, we can take the derivative of R with respect to x, set it equal to zero, and solve for x. After solving, we find that x = 15, which means the ticket price that will result in the greatest revenue is $15.

4) For the fourth question, we have a 240km trip to the cottage. The Nakamura family reduced their average speed by 20 km/h, which increased their journey time by 1 hour. We want to find the slower average speed.

Let's denote the original average speed as 'x' km/h and the slower average speed as 'y' km/h. We know that the distance traveled is the same (240km). The time it takes at the original speed is given by 240/x hours, and the time it takes at the slower speed is 240/y hours.

Since the slower speed increased the journey time by 1 hour, we can set up the equation:

240/x + 1 = 240/y

To solve for y in terms of x, we can rearrange the equation:

1 = 240/y - 240/x

Simplifying further, we get:

1 = (240x - 240y) / (xy)

Now, let's solve for y:

xy = 240x - 240y

xy + 240y = 240x

y(x + 240) = 240x

y = 240x / (x + 240)

To find the slower average speed, we can substitute different values for x and calculate y. When x = 80 km/h, y ≈ 60 km/h. Therefore, the slower average speed is 60 km/h.

5) For the fifth question, we have 3 cubes stacked on a desk, where the edges of the cubes are consecutive odd integers. We want to find the lengths of the sides of the cubes.

Let's denote the lengths of the sides as 'x', 'x + 2', and 'x + 4'. The total exposed surface area of the cubes is 281 cm^2.

The surface area of a cube is given by: A = 6 * (side length)^2

We can set up the equation:

6(x^2 + (x + 2)^2 + (x + 4)^2) = 281

Simplifying and solving the quadratic equation, we find that x = 3. Therefore, the lengths of the sides of the cubes are 3cm, 5cm, and 7cm.

I hope these explanations help you understand how to approach these math problems! If you have any further questions, feel free to ask. Good luck with your assignment!