Mathematical Skills IV Honor

Science & Health

Requirements

1. Have the Mathematical Skills III Honor.

   Answer: You must have previously completed the Mathematical Skills III Honor, demonstrating mastery of the progressive content: simple equations, fractions, introductory percentages, basic geometry, and operations with decimals and roots — an essential foundation before advancing to quadratic equations and functions in this Honor IV. — The Mathematical Skills series (I, II, III, and IV) follows a progressive curriculum from the final grades of Brazilian elementary school to high school — it was designed by the South American Division in 2012 to align with the BNCC.

2. Present a bibliographic report with at least five figures who contributed to the development of mathematics throughout the history of humanity.

   Answer: Include: Pythagoras (6th century BC, theorem of right triangles); Euclid (3rd century BC, axiomatic geometry in 'The Elements'); Archimedes (287-212 BC, calculation of the number π and principles of physics); Al-Khwarizmi (9th century AD, founder of modern algebra and of the algorithm); Isaac Newton (1643-1727, differential and integral calculus). — Al-Khwarizmi is the origin of the words 'algebra' (from the book Al-jabr) and 'algorithm' (a Latinization of his name) — the Persian-Arab scholar is the father of the modern mathematics we know.

3. Work out and present the calculations of the following equations:
   • 5x² - 3x - 2 = 0
   • 3x² + 55 = 0
   • x² - 10x + 25 = 0

   Answer: a) 5x²-3x-2=0: Δ=b²-4ac=9+40=49; x=(3±7)/10; x'=1 and x''=-2/5. b) 3x²+55=0: 3x²=-55; x²=-55/3 (negative); NO REAL SOLUTION since Δ<0. c) x²-10x+25=0: Δ=100-100=0; x=10/2=5 (double root, perfect square). — Equation (c) is a perfect square trinomial: x²-10x+25 = (x-5)² — characterized by Δ=0 and a single root. Equation (b) has coefficient a>0 with c>0 and b=0, always without real roots.

4. Present and work out the percentage calculations of the following problems:
   • 3% of 450
   • 25% of 1440
   • 30% of 2500

   Answer: a) 3% of 450 = (3/100) × 450 = 13.5. b) 25% of 1440 = (25/100) × 1440 = 360 (or 1/4 of 1440). c) 30% of 2500 = (30/100) × 2500 = 750. To calculate x% of N, multiply N by (x/100), or divide N by 100 and multiply by x. — The practical rule '1% of N = N/100' speeds up the calculations: 1% of 450 = 4.5, so 3% = 13.5; 1% of 2500 = 25, so 30% = 750 — a mental method used in commerce and finance.

5. Present in writing three practical situations where we use percentage in our daily lives.

   Answer: Three practical uses: (1) store discounts — a R$80 shirt with a 20% discount costs R$64 (R$80 × 0.80); (2) credit card interest — a debit balance of R$1000 with 8% monthly interest generates an extra R$80 per month; (3) taxes on products — an 18% ICMS on goods worth R$500 adds R$90 to the final price. — The revolving credit card interest rate in Brazil reached 437% per year in 2023 (Central Bank), making financial literacy about percentages crucial — Law 14,690/2023 capped interest at 100% per year within up to 60 days.

6. Present and work out three practical situations of everyday problems in which we use the quadratic equation.

   Answer: Three situations: (1) a rectangular lot whose area is 200m² and whose length is twice the width — set up 2x²=200, x=10m width and 20m length; (2) free-fall motion h=½gt² — calculate the time of fall; (3) maximum profit in a company whose function L(x)=-x²+100x allows finding the vertex of the parabola. — The equation 2x²=200 from the lot problem has an elegant solution: x²=100, x=±10. We take only the positive value (a measurement cannot be negative) — discarding a root is common practice in real physical problems.

7. Solve and present the working of the following functions: Given f(x) = x - 3 and g(x) = -3x + 4, determine:
   • f(f(0))
   • f(f(1)) + g(f(3))

   Answer: f(x)=x-3, g(x)=-3x+4. (a) f(f(0)): first f(0)=0-3=-3; then f(-3)=-3-3=-6. So f(f(0))=-6. (b) f(f(1))+g(f(3)): f(1)=1-3=-2; f(-2)=-2-3=-5. For g(f(3)): f(3)=3-3=0; g(0)=-3(0)+4=4. Total: -5+4=-1. — A composite function f(g(x)) means applying g first and then f to the result — order matters because f∘g is generally different from g∘f, a property called the non-commutativity of composition.

8. Represent the following functions on the Cartesian graph:
   • y = 3x - 1
   • f(x) = 2x + 3

   Answer: For y=3x-1: a straight line with slope 3 (rises 3 units for each 1 horizontal) and y-intercept -1; passing through the points (0,-1) and (1,2). For f(x)=2x+3: a line with slope 2 and y-intercept 3; passing through (0,3) and (1,5). Plot both on the Cartesian plane with x and y axes, connecting the calculated points with a ruler. — Every function of the type y=ax+b is a straight line — the coefficient a is the slope (rise/run) and b is where the line crosses the y-axis. Known as the slope-intercept form of the line's equation, it is essential in calculus and analytic geometry.

9. Demonstrate the ability to solve problems involving circles, such as calculating the circumference and the area using the formula for each. Present two examples of each.

   Answer: Formulas: Circumference C=2πr (or C=πd where d=diameter). Area A=πr². Circumference examples: (1) r=5cm: C=2π(5)=10π≈31.42cm; (2) r=10cm: C=20π≈62.83cm. Area examples: (1) r=5cm: A=π(5²)=25π≈78.54cm²; (2) r=10cm: A=100π≈314.16cm². — The number π (pi) ≈ 3.14159 is the constant ratio between the circumference and the diameter of any circle — Archimedes calculated π for the first time in 250 BC using inscribed and circumscribed polygons.

10. Demonstrate the ability to calculate the area of regular polygons, such as a hexagon inscribed in a circle, the surface area of the cylinder, the volume of the prism, and the volume of the pyramid.

    Answer: Inscribed hexagon (side=radius): A=(3√3/2)·l². Surface area of the cylinder: A=2πr²+2πrh (two caps + lateral). Volume of the prism: V=A_base·h. Volume of the pyramid: V=(1/3)·A_base·h. Cylinder example r=3, h=10: A=2π(9)+2π(3)(10)=18π+60π=78π≈245cm². — The pyramid has 1/3 the volume of the prism with the same base and height — this factor of 1/3 was demonstrated by Eudoxus in the 4th century BC using the method of exhaustion, preceding Newton's integral calculus.

11. In our daily lives, we deal with interest rates all the time. Demonstrate the ability to solve the two most common interest situations.
    • Luciana made an investment of R$ 200.00 at simple interest of 2% per month. How much will she have, in total, after 8 months of investment?
    • Davi took out a bank loan of R$ 3,000.00 and will pay it off in 6 months at a compound interest rate of 1.5% per month. Calculate the total he will have to pay the bank after these 6 months.

    Answer: Simple interest (Luciana): J=C·i·t = 200·0.02·8 = R$32; final total = 200+32 = R$232. Compound interest (Davi): M=C·(1+i)^t = 3,000·(1.015)^6 ≈ 3,000·1.0934 ≈ R$3,280.18. Difference: simple interest grows linearly, compound interest grows exponentially because it is applied to the accumulated balance. — The difference between simple and compound interest is that compound interest incorporates 'interest on interest' — over long periods the difference is enormous, which is why the Central Bank has required the disclosure of the CET (Total Effective Cost) in every financing arrangement since 2008.