<p>Name______</p><p>Math Assessment Do not use any resource (computer, textbook, human, etc.) besides a calculator. Show as much work as possible and minimize your use of a calculator. Indicate when you use a calculator. </p><p>Unit Conversion I expect you to be familiar with this, but don’t worry if you are rusty since it is relatively straightforward to learn and we will spend a little class time on this.</p><p>Given: 1.00 inch = 2.54 cm, 1.00 m = 100 cm</p><p>Convert to inches: 2.5 m</p><p>Solution: 2.5 m (100 cm/1 m) (1 inch/2.54 cm) = 98.4 inches</p><p>Convert to square centimeters: 1.75 m2</p><p>Solution: 1.75 m2*(100 cm/1 m)2 = 17500 cm2</p><p>Algebra You must be able to do algebra to succeed in this class. We will devote little class time to this subject. If you are not able to do this work without help at this time, you should consider delaying your enrollment in this class.</p><p>1) Given: -3x + x2 = 7</p><p>Solve the above equation for x.</p><p>Solution: Shove everything to one side and use the quadratic formula or a calculator. x2 + -3x + -7 = 0 x = (3 ± sqrt[(-3)2 – 4*1*-7)/2 = 4.54, -1.54</p><p>2) Given: 2x + 4 = 3y x + y = 5</p><p>Solve the above pair of equations for x and y.</p><p>Solution: There are several methods to solve this, but I usually use the substitution method. You can of course use your calculator solver. x = 5 – y (this is the second equation solved for x) 2(5 – y) + 4 = 3y (the second equation plugged into the first) 10 – 2y + 4 = 3y 14 = 5y y = 14/5</p><p>Now plug the above value into the second equation solved for x. x = 5 – 14/5 x = 11/5</p><p>3) Given: ab + c = d cd = a/b</p><p>Solve the above pair of equations for a and b in terms of c and d. Simplify your expressions as much as possible.</p><p>Solution: Use the exact same method as in problem #2. a = bcd (the second equation solved for a) (bcd)b + c = d (the second equation plugged into the first) b2cd + c = d b2cd = d – c b2 = (d – c)/(cd) b2 = 1/c – 1/d (this simplifies the expression) b = sqrt(1/c – 1/d)</p><p>Now plug the above solution into the second equation solved for a a = sqrt(1/c – 1/d)cd Vector Arithmetic You must be able to do vector arithmetic to succeed in this class. We will devote some class time to this subject, but it should be as a review rather than as new material. If you are not able to do this work on your own after the first week, you should consider delaying your enrollment in this class.</p><p>Given: A = 3î - 4ĵ, B = -2î + 5ĵ, C = (200, 50°), and D = (60, 20°) Note that some texts use different notation for vectors. Vector A could be written as (3, -4), 3x – 4y, or Ax = 3 and Ay = -4. Vector C could be written as C = 200 and θ = 50°.</p><p>1) A + B = (3î - 4ĵ) + (-2î + 5ĵ) = (3î + -2î) + (-4ĵ + 5ĵ) = 1î + 1ĵ</p><p>2) Convert A to polar coordinates (magnitude and direction) A = sqrt(32 + (-4)2) = 5 θ = tan-1(-4/3) = -53° A = (5, -53°)</p><p>3) Convert C to Cartesian coordinates (x, y) Cx = C*cos(θ) = 200*cos(50°) = 128 Cy = C*sin(θ) = 200*sin(50°) = 153 C = (128, 153)</p><p>4) (dot product) A·B = AxBx + AyBy = 3*-2 + -4*5 = -26</p><p>5) (dot product) C·D = C*D*cosφ = 200*60*cos(50° - 20°) = 10392</p><p>6) C + D = Dx = D*cos(θ) = 60*cos(20°) = 56 Dy = D*sin(θ) = 60*sin(20°) = 20 Cx + Dx = 128 + 56 = 184 Cy + Dy = 153 + 20 = 173 C + D = (184, 173) (Note: it is common practice to convert back to polar as follows) sqrt(1842 + 1732) = 253 θ = tan-1(173/184) = 43° C + D = (253, 43°)</p>