course Mth 158
......!!!!!!!!...................................
......!!!!!!!!...................................
RESPONSE --> R = xp a) R = -20p + 500(p) R = -20p^2 + 500p b)R = -20(20)^2 + 500(20) R = -8000 + 10000 R = $2000
.................................................
......!!!!!!!!...................................
17:39:14 ** revenue = demand * price = x * p = (-20 p + 500) * p = -20 p^2 + 500 p If price = 24 then we get R = -20 * 24^2 + 500 * 24 = 480. **
......!!!!!!!!...................................
RESPONSE --> .
.................................................
......!!!!!!!!...................................
17:42:01 query 3.6.10. P = (x, y) on y = x^2 - 8. Give your expression for the distance d from P to (0, -1)
......!!!!!!!!...................................
RESPONSE --> d = sqrt((x - 0)^2 + (y - -1)^2) d = sqrt(x^2 + (x^2 - 8) + 1)^2 d= sqrt(x^4 - 13x^2 + 49)
.................................................
......!!!!!!!!...................................
17:42:07 ** P = (x, y) is of the form (x, x^2 - 8). So the distance from P to (0, -1) is sqrt( (0 - x)^2 + (-1 - (x^2-8))^2) = sqrt(x^2 + (-7-x^2)^2) = sqrt( x^2 + 49 - 14 x^2 + x^4) = sqrt( x^4 - 13 x^2 + 49). **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
17:43:21 What are the values of d for x=0 and x = -1?
......!!!!!!!!...................................
RESPONSE --> d = sqrt(0^4 - 13(0)^2 + 49 = sqrt(49) = 7 d = sqrt(-1^4 - 13(-1)^2 + 49 = sqrt(37)
.................................................
......!!!!!!!!...................................
17:45:11 ** If x = 0 we have sqrt( x^4 - 13 x^2 + 49) = sqrt(0^4 - 13 * 0 + 49) = sqrt(49) = 7. If x = -1 we have sqrt( x^4 - 13 x^2 + 49) = sqrt((-1)^4 - 13 * (-1) + 49) = sqrt( 64) = 8. Note that these results are the distances from the x = 0 and x = 1 points of the graph of y = x^2 - 8 to the point (0, -1). You should have a sketch of the function and you should vertify that these distances make sense. **
......!!!!!!!!...................................
RESPONSE --> How did you get 64
.................................................
......!!!!!!!!...................................
17:46:37 query 3.6. 18 (was and remains 3.6.18). Circle inscribed in square. What is the expression for area A as a function of the radius r of the circle?
......!!!!!!!!...................................
RESPONSE --> Area of circle is pi r^2 One side of square will be 2r Area of square is (2r)^2 = 4r Area of square minus area of circle is 4r^2 - pi r^2
.................................................
......!!!!!!!!...................................
17:46:40 ** A circle inscribed in a square touches the square at the midpoint of each of the square's edges; the circle is inside the square and its center coincides with the center of the square. A diameter of the circle is equal in length to the side of the square. If the circle has radius r then the square has sides of length 2 r and its area is (2r)^2 = 4 r^2. The area of the circle is pi r^2. So the area of the square which is not covered by the circle is 4 r^2 - pi r^2 = (4 - pi) r^2. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
17:46:53 What is the expression for perimeter p as a function of the radius r of the circle?
......!!!!!!!!...................................
RESPONSE --> 4(2r) = 8r
.................................................
......!!!!!!!!...................................
17:46:56 ** The perimeter of the square is 4 times the length of a side which is 4 * 2r = 8r. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
17:48:18 query 3.6.27 (was 3.6.30). one car 2 miles south of intersection at 30 mph, other 3 miles east at 40 mph Give your expression for the distance d between the cars as a function of time.
......!!!!!!!!...................................
RESPONSE --> d = sqrt(2 - 30t)^2 + (3 - 40t)^2 d = sqrt(4 - 120t + 900t^2 + 9 - 240t + 1600t^2) d = sqrt(2500t^2 - 360t + 13)
.................................................
......!!!!!!!!...................................
17:48:21 ** At time t the position of one car is 2 miles south, increasing at 30 mph, so its position function is 2 + 30 t. The position function of the other is 3 + 40 t. If these are the x and the y coordinates of the position then the distance between the cars is distance = sqrt(x^2 + y^2) = sqrt( (2 + 30 t)^2 + (3 + 40t)^2 ) = sqrt( 4 + 120 t + 900 t^2 + 9 + 240 t + 1600 t^2) = sqrt( 2500 t^2 + 360 t + 13). **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
"