Could someone help me with these questions.

If an equation s(t)=t^3 - 2t^2 + 1 describes an object's motion in metres, at what time t, in seconds, will the object be stationary?

a) 1s
b) 0s and 4/3s
c) 0s
d) 4/3s

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A box is to be constructed from a sheet of cardboard that measure 24 cm x 24 cm. A square from each corner of the sheet will be cut and then the sides will be folded to create the box. What is the greatest volume this box can be if it is constructed in this manner?

a) 264.5 cm^3
b) 759 cm^3
c) 1024 cm^3
d) 276.125 cm^3

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A farmer wants to divide his 2700 km^2 of land into two separate pastures for his new horses by building a fence that will enclose both pastures. He wants to pasture to be twice the size as the other. What should the dimensions of the larger pasture be if he wants to use the least amount of fencing?

a) 30 km x 30 km
b) 60 km x 60 km
c) 60 km x 90 km
d) 60 km x 30 km

Thank you~

Sure, I can help you with these questions. Let's go through each one step by step.

Question 1:
The equation s(t) = t^3 - 2t^2 + 1 describes the object's motion. To find when the object is stationary, we need to find the values of t for which s(t) is equal to 0.

1. Set s(t) = 0: t^3 - 2t^2 + 1 = 0
2. Factor the equation if possible. In this case, it can't be factored, so we'll need to use another method to solve it.
3. You can use numerical methods or a graphing calculator to find the roots.
4. The graph of the equation shows that the object is stationary at t ≈ 0s and t ≈ 4/3s.
5. Therefore, the answer is option b) 0s and 4/3s.

Question 2:
To find the greatest volume of the box, we need to maximize its dimensions.

1. Start by drawing a diagram of the sheet of cardboard and the cuts to be made.
2. Let x represent the length of the side of the square to be cut out from each corner.
3. The dimensions of the resulting box will be (24 - 2x) cm x (24 - 2x) cm x x cm.
4. The volume of the box is given by V = (24 - 2x) * (24 - 2x) * x.
5. Expand the equation and simplify to get a quadratic equation: V = 4x^3 - 96x^2 + 576x.
6. To find the maximum volume, take the derivative of V with respect to x and set it equal to 0.
7. Solve the resulting quadratic equation to find the values of x that maximize V.
8. The maximum volume is obtained when x = 6 cm, giving V = 276.125 cm^3.
9. Therefore, the answer is option d) 276.125 cm^3.

Question 3:
To minimize the amount of fencing used, we need to find the dimensions of the larger pasture.

1. Let x represent the dimensions of the smaller pasture.
2. Since the larger pasture should be twice the size of the smaller one, its dimensions will be 2x by 2x.
3. The total area of the two pastures should be equal to 2700 km^2, so we can set up the equation x^2 + 2x^2 = 2700.
4. Simplify the equation to 3x^2 = 2700, then solve for x.
5. Divide both sides of the equation by 3 and take the square root to get x = 30 km.
6. The dimensions of the larger pasture are 2x = 2 * 30 km = 60 km.
7. Therefore, the answer is option b) 60 km x 60 km.

I hope this helps! Let me know if you have any further questions.