To prove: The polynomials P x = 3 x 4 − 5 x 3 + x 2 − 3 x + 5 and Q x = 3 x − 5 x + 1 x − 3 x + 5 are the same by expanding Q , and then calculate the values of P 2 and Q 2 to determine the easier method. The polynomials Q x and P x are the same ¯ ; the values of P 2 and Q 2 is 11. Property used: Distributive property: a b + c = a b + a c Calculation: Use the distributive property and expand the polynomial Q x as follows. Q x = 3 x − 5 x + 1 x − 3 x + 5 = 3 x − 5 x + 1 x 2 − 3 x + 5 = 3 x − 5 x 3 + x 2 − 3 x + 5 = 3 x 4 − 5 x 3 + x 2 − 3 x + 5 Observe that this is same as P x = 3 x 4 − 5 x 3 + x 2 − 3 x + 5 . The values of Q 2 and P 2 are computed as follows. Calculate the value of Q 2 by substituting 2 for x in Q x = 3 x − 5 x + 1 x − 3 x + 5 . Q 2 = 3 × 2 − 5 2 + 1 2 − 3 2 + 5 = 6 − 5 2 + 1 2 − 3 2 + 5 = 1 2 + 1 2 − 3 2 + 5 = 2 + 1 2 − 3 2 + 5 Simplify further as follows. Q 2 = 3 2 − 3 2 + 5 = 6 − 3 2 + 5 = 3 ⋅ 2 + 5 = 11 Thus, the value of Q 2 is 11. Now, calculate the value of P 2 by substituting 2 for x in P x = 3 x 4 − 5 x 3 + x 2 − 3 x + 5 . P 2 = 3 × 2 4 − 5 × 2 3 + 2 2 − 3 × 2 + 5 = 3 × 16 − 5 × 8 + 4 − 6 + 5 = 48 − 40 + − 2 + 5 = 11 Thus, the value polynomial P x is 11. From the above calculations, observe that finding the value of the polynomial in the expanded form is easier than that in the nested form. Thus, the polynomials Q x and P x are the same ¯ ; the values of P 2 and Q 2 is 11. To write: The polynomial R x = x 5 − 2 x 4 + 3 x 3 − 2 x 2 + 3 x + 4 in nested form and find the value of R 3 by substitution and by using synthetic division. The polynomial R x in nested form is x − 2 x + 3 x − 2 x + 3 x + 4 ¯ and the value of R 3 is 157 ¯ . Write R x = x 5 − 2 x 4 + 3 x 3 − 2 x 2 + 3 x + 4 in nested form as follows. R x = x 4 − 2 x 3 + 3 x 2 − 2 x + 3 x + 4 ∵ By distributive property = x 3 − 2 x 2 + 3 x − 2 x + 3 x + 4 = x 2 − 2 x + 3 x − 2 x + 3 x + 4 = x − 2 x + 3 x − 2 x + 3 x + 4 Therefore, the polynomial R x in nested form is x − 2 x + 3 x − 2 x + 3 x + 4 . Compute the value of R 3 by substituting 3 for x in R x = x − 2 x + 3 x − 2 x + 3 x + 4 . R 3 = 3 − 2 3 + 3 3 − 2 3 + 3 3 + 4 = 1 3 + 3 3 − 2 3 + 3 3 + 4 = 3 + 3 3 − 2 3 + 3 3 + 4 = 6 3 − 2 3 + 3 3 + 4 Further solve the equation, R 3 = 16 3 + 3 3 + 4 = 48 + 3 3 + 4 = 51 × 3 + 4 = 153 + 4 = 157 Thus, the polynomial R x in nested form is x − 2 x + 3 x − 2 x + 3 x + 4 ¯ and the value of R 3 is 157 ¯ . Use synthetic division to divide R x = x 5 − 2 x 4 + 3 x 3 − 2 x 2 + 3 x + 4 by x − 3 as follows: Abbreviate the dividend R x = x 5 − 2 x 4 + 3 x 3 − 2 x 2 + 3 x + 4 by writing the coefficients 1 , − 2 , 3 , − 2 , 3 , 4 and write 3 instead of the divisor x − 3 . Bring down the first number and multiply by the divisor. Write the result in the middle row, below the second number. Then add the second number and the number written below. Repeat this process until the table is complete. Note that, the last number obtained will be the remainder. 3 1 − 2 3 − 2 3 4 3 3 18 48 153 1 1 6 16 51 157 Thus, the remainder is 157. Therefore, the evaluation of R 3 with nested equation follows the same arithmetic steps as calculating the value o f the polynomial using synthetic division.

The cup on the 9th hole of a golf course is located dead center in the middle of a circular green which is 40 feet in radius. Your ball is located as in the picture below. The ball follows a straight line path and exits the green at the right-most edge. Assume the ball travels 8 ft/sec. Introduce coordinates so that the cup is the origin of an xy-coordinate system and start by writing down the equations of the circle and the linear path of the ball. Provide numerical answers below with two decimal places of accuracy. 50 feet green ball 40 feet 9 cup ball path rough (a) The x-coordinate of the position where the ball enters the green will be (b) The ball will exit the green exactly seconds after it is hit. (c) Suppose that L is a line tangent to the boundary of the golf green and parallel to the path of the ball. Let Q be the point where the line is tangent to the circle. Notice that there are two possible positions for Q. Find the possible x-coordinates of Q: smallest x-coordinate =…

Draw the unit circle and plot the point P=(8,2). Observe there are TWO lines tangent to the circle passing through the point P. Answer the questions below with 3 decimal places of accuracy. P L1 L (a) The line L₁ is tangent to the unit circle at the point (b) The tangent line L₁ has equation: X + (c) The line L₂ is tangent to the unit circle at the point ( (d) The tangent line 42 has equation: y= x + ).

What is a solution to a differential equation? We said that a differential equation is an equation that describes the derivative, or derivatives, of a function that is unknown to us. By a solution to a differential equation, we mean simply a function that satisfies this description. 2. Here is a differential equation which describes an unknown position function s(t): ds dt 318 4t+1, ds (a) To check that s(t) = 2t2 + t is a solution to this differential equation, calculate you really do get 4t +1. and check that dt' (b) Is s(t) = 2t2 +++ 4 also a solution to this differential equation? (c) Is s(t)=2t2 + 3t also a solution to this differential equation? ds 1 dt (d) To find all possible solutions, start with the differential equation = 4t + 1, then move dt to the right side of the equation by multiplying, and then integrate both sides. What do you get? (e) Does this differential equation have a unique solution, or an infinite family of solutions?

Which degenerate conic is formed when a double cone is sliced through the apex by a plane parallel to the slant edge of the cone?

1/ Solve the following: 1 x + X + cos(3X) -75 -1 2 2 (5+1) e 5² + 5 + 1 3 L -1 1 5² (5²+1) 1 5(5-5)

I need expert handwritten solution.to this integral