Equations, expressions, and inequalitie #11 - Questions and Answers

Question 1 Simplify 2x+x. Answer 2x + x = 2 x x + 1 x x = x x (2+1) = 3x Question 2 Write an algebraic expression for: 6 more than a number c. Answer more than =+ ⇒6+c Result: 6+c Question 3 The solution set for polynomial inequality 2x² + 5x - 3 ≤ 0 using a graphing utility. Answer 1 Polynomial inequality is an inequality that can be put in any of the following forms: f(x)<0, f(x)>0, f(x) ≤ 0, or f(x) ≥ 0 where f is any polynomial function. A graph technique is used to visualize the solutions of polynomial inequalities. For this, use x-intercepts of given polynomial function f as boundary points that divide the real number line into intervals. On each interval, the graph of f is either above the x-axis [f(x)>0] or below the x-axis [f(x) <0]. This fact gives reasons for x-intercept to play fundamental role in solving polynomial inequalities. The x-intercepts can be found by solving the equation f(x) = 0. Plot the graph of 2x² + 5x - 3 = 0 using the following steps of the Ti-83 calculator: Step 1: Press the [Y =] key: the equations for y will appear. Step 2: Enter the equations in Y₁. Here, Y₁ = 2X²+5X-3. Step 3: Go to the leftmost side of the line, that is left of Y₁, and press Enter until you get the sign of >. Step 4: Press the [Trace] or [GRAPH] key to plot the graph. Not, look for the points on the x-axis and mention the solution set, that is, [-3,1/2]. Conclusion: The solution set for the given polynomial inequality 2x²+5x-3 ≤ 0 is [-3,1/2]. Answer 2 Step 1 2x²+5x-3 ≤ 0 To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation ax² + bx + c = a(x - x₁)(x - x₂), where x₁ and x₂ are the solution of the quadratic equation ax² + bx + c = 0 2x²+5x-3=0 All equations of the form ax² + bx + c = 0 can be solved using the quadratic formula: x= (-b±√(b²-4ac))/2a Substitute 2 for a, 5 for b, and -3 for c in the quadratic formula. x= (-5±√(5²-4×2(-3)))/(2×2) Do the calculations. x = (-5±7)/4 Solve the equation x = (-5±7)/4 when + is plus and when ± is minus x = 1/2 x = -3 Rewrite the inequality by using the obtained solutions. 2(x-1/2)(x+3)≤0 For the product to be ≤ 0, one of the values x-1/2 and x + 3 has to be ≥ 0 and the other has to be ≤ 0. Consider the case when x-1/2 ≥ 0 and x +3≤0 x-1/2 ≥ 0 x + 3 ≤ 0 This is false for any x. Step 2 Consider the case when x-1/2 ≤ 0 and x + 3 ≥ 0 x + 3 ≥ 0 x -1/2 ≤ 0 The solution satisfying both inequalities is x ∈ [-3, 1/2] The final solution is the union of the obtained solutions. x ∈ [-3, 1/2] Question 4 Find the sum of 2 cubed and 3 squared. Answer 2³+3² =8+9 =17 Question 5 A dockworker applies a constant horizontal force of 80.0 N to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves 11.0 m in 5.00 s. What is the mass of the block of ice? If the worker stops pushing at the end of 5.00 s, how far does the block move in the next 5.00 s? Answer 1 (1) x-xo = 11.0m t = 5.00s Vox = 0. x-xo = Voxt + 1/2axt² PLAN: the acceleration, then multiply by ∑Fx = max Let +x be the direction of the force. ∑Fx = 80.0N ax = 2(x-xo)/t² = 2 (11.0m)/ (5.00s)² = 0.880 m/s² m = ∑Fx / ax = 80.0N / 0.880 m/s² = 90.9kg (2) aₓ = 0 and vₓ is constant. After the first 5.0 s, vₓ = v₀ₓ + aₓt = (0.880 m/s²)(5.00s) = 4.40 m/s The motion of the next 5s (a = 0) will now move at this initial speed x - xo = voxt + 1/2 axt² = (4.40 m/s)(5.00s) = 22.0m For (2), first calculate the speed at the end of the period (5s) of applied force (accelerated kinematics) Utilize this finishing velocity as the initial (and constant) value for the second component of the motion then. Answer 2 x = xo + voxt + 1/2 axt² Or aₓ = 2x/t² = 2(11m)/(5s)² = 0.88 m/s² [image of diagram] Answer 3 Given Force = 80 N Initial velocity (u) = 0 Distance moved 11 m in 5s s = ut + 1/2 at² 11 = 0 + 1/2 a x 25 22 = ax 25 a = 22/25 a = 0.88 m/s² and force = mass x acceleration 80 = m x 0.88 m = 90.9kg Block acquires velocity at the conclusion of 5 seconds. v = u + at v = 0 + 0.88 x 5 = 4.4 m/s Distance moved in next 5 sec s = vt + 0 s = 4.4 x 5 = 22m Answer 4 Step 1) To find the mass of the block of ice, we can use Newton's second law of motion: Force = mass x acceleration Since the block moves with a constant velocity, its acceleration is zero. Therefore, the applied force is equal to the force of inertia: Force = mass x acceleration = mass x 0 = 0 Since the force applied by the dockworker is the only horizontal force acting on the block, the mass of the block of ice is 0 kg. Step 2) If the worker stops pushing at the end of 5.00 s, the block will continue to move with its current velocity due to its inertia. Since there are no horizontal forces acting on the block, it will move with a constant velocity in the next 5.00 s. Therefore, the block will move the same distance as it did in the first 5.00 s, which is 11.0 m. Hence, the block will move 11.0 m in the next 5.00 s. Answer 5 The mass of the block of ice is approximately 90.9 kg. If the worker stops pushing at the end of 5.00 s, the block will move approximately 11.0 m in the next 5.00 s. Explanation: To solve the given problem, let's use the following equations of motion: v = u + at s = ut + 1/2 at² v² = u² + 2as Where: s represents the displacement (distance traveled) u represents the initial velocity v represents the final velocity a represents the acceleration t represents the time Given: The horizontal force applied, F = 80.0 N The displacement, s = 11.0 m The time, t = 5.00 s To find the mass of the block of ice, we'll use Newton's second law of motion, F = ma, where m is the mass and a is the acceleration. Since the frictional force is negligible, the only force acting on the block is the applied force. Therefore, F = ma. We can rearrange the equation to solve for mass: m = F/a Now, let's calculate the acceleration using the given displacement and time. Using equation 2, s = ut + 1/2 at², we can rearrange it to solve for acceleration: a = 2(s-ut)/t² Substituting the given values into the equation, we get: a = 2(11.0 - 0 x 5.00) / (5.00)² Simplifying the equation: a = 22.0 / 25.0 Now, we can substitute the acceleration into the equation for mass: m = 80.0 / (22.0/25.0) Simplifying the equation: m = (80.0 x 25.0) / 22.0 m ≈ 90.9 kg Therefore, the mass of the block of ice is approximately 90.9 kg. If the worker stops pushing at the end of 5.00 s, the block will continue to move with the initial velocity it acquired during the first 5.00 s. So, the final velocity, v, remains the same. To find the distance traveled in the next 5.00 s, we'll use equation 2: s = ut + 1/2 at² Since the block starts from rest, the initial velocity, u, is 0. The time, t, is 5.00 s, and we already know the acceleration, a, from the previous calculation. Substituting the values into the equation: s = 0 x 5.00 + 1/2 x (22.0/25.0) x (5.00)² Simplifying the equation: s = 22.0/25.0 x 12.5 s ≈ 11.0 m Therefore, the block will move approximately 11.0 m in the next 5.00 s.