#1**0 **

AnthonywardMar 22, 2023

#1**0 **

AnthonywardMar 21, 2023

#1**0 **

(a) To describe the graph of the parametric equations, we can eliminate the parameter t to obtain an equation in terms of x and y. From the given equations, we have:

sin pit = x - 1 + cos pit

cos pit = (y - 3sin pi*t)/2

Substituting the first equation into the second equation, we get:

cos pit = (y - 3(x - 1 + cos pit))/2

2cos pit = y - 3x + 3

cos pit = (y - 3x + 3)/2

Substituting this into the first equation, we get:

sin pit = x - 1 + (y - 3x + 3)/2

2sin pit = 2x - 2 + y - 3x + 3

sin pit = x + y - 1

Therefore, the graph of the parametric equations is the set of all points (x, y) that satisfy the equation x + y - 1 = sin pi*t, where t ranges over all real numbers.

(b) To describe the motion of the particle as t ranges from 0 to 2, we can evaluate x and y at t = 0 and t = 2, and plot the resulting points.

At t = 0, we have:

x = 1 + sin 0 - cos 0 = 1

y = 3sin 0 + 2cos 0 = 2

At t = 2, we have:

x = 1 + sin pi2 - cos pi2 = 1

y = 3sin pi2 + 2cos pi2 = -1

Therefore, the particle moves horizontally along the line y = 2, then returns to its starting point by moving horizontally along the line y = -1.

(c) To find a parametrization that matches the graph of part (b) but with a different motion, we can use a different function for x or y. For example, we can use:

x = 1 + sin pit

y = 2cos pi*t

This parametrization produces the same graph as the original parametrization for t ranging from 0 to 2, but the motion of the particle is different. In this case, the particle moves along an ellipse centered at (1, 0), starting from the point (2, 0) and ending at the point (0, 0). HEB Partner

AnthonywardMar 17, 2023

#3**0 **

The area of a sector bounded by a 180° arc of a circle with a radius of 9 inches is:

$$\frac{1}{2} \times \pi \times r^2 \times \frac{\theta}{360} = \frac{1}{2} \times \pi \times 9^2 \times \frac{180}{360} = \frac{1}{2} \times \pi \times 81 = \frac{81}{2} \pi \approx 127.23\text{ in}^2$$

Therefore, the area of the sector is approximately 127.23 square inches. SkyWard Alpine Login

AnthonywardMar 16, 2023

#1**-1 **

AnthonywardMar 14, 2023

#1**0 **

Let's call the twins A and B, and the other three children C, D, and E. We can approach this problem by breaking it down into cases:

Case 1: A and B each get 1 piece of candy

In this case, we have 8 pieces of candy left to distribute to 5 children (A, B, C, D, E) without any restrictions. This is equivalent to distributing 8 identical pieces of candy to 5 children, which can be done in (8 + 5 - 1) choose (5 - 1) = 462 ways using stars and bars.

Case 2: A and B each get 2 pieces of candy

In this case, we have 6 pieces of candy left to distribute to 3 children (C, D, E) without any restrictions. This is equivalent to distributing 6 identical pieces of candy to 3 children, which can be done in (6 + 3 - 1) choose (3 - 1) = 21 ways using stars and bars.

So the total number of ways to distribute the candy is 462 + 21 = 483. My Sutter Online Login

AnthonywardMar 13, 2023

#2**0 **

AnthonywardMar 11, 2023