Integrated Math 10th Grade 1st Semester Exhaustive Study Note

Coordinate Geometry and Lines

• Analysis of Segment AB with Endpoints $A(-5, 4)$ and $B(3, -6)$:     - Finding the Perpendicular Bisector:         - The perpendicular bisector of a segment passes through its midpoint and has a slope that is the negative reciprocal of the segment's slope.         - Step 1: Find the Midpoint ($M$): $M = \left(\frac{-5+3}{2}, \frac{4+(-6)}{2}\right) = \left(\frac{-2}{2}, \frac{-2}{2}\right) = (-1, -1)$.         - Step 2: Find the Slope ($m$) of AB: $m = \frac{-6-4}{3-(-5)} = \frac{-10}{8} = -\frac{5}{4}$.         - Step 3: Find the Perpendicular Slope ($m_{\perp}$): Since the product of perpendicular slopes is $-1$, $m_{\perp} = \frac{4}{5}$.         - Step 4: Slope-Intercept Equation ($y = mx + b$): Use the midpoint $(-1, -1)$ and $m_{\perp} = \frac{4}{5}$.           $y - (-1) = \frac{4}{5}(x - (-1))$           $y + 1 = \frac{4}{5}x + \frac{4}{5}$           $y = \frac{4}{5}x - \frac{1}{5}$.     - Algebraic Verification of Point $C(-6, -5)$:         - To show algebraically if point $C$ is on the perpendicular bisector, substitute its coordinates into the equation: $-5 = \frac{4}{5}(-6) - \frac{1}{5}$.         - $-5 = -\frac{24}{5} - \frac{1}{5}$         - $-5 = -\frac{25}{5}$         - $-5 = -5$. Since the equation holds true, $C$ lies on the perpendicular bisector.     - Distance Properties (AC and BC):         - Calculate the distance between point $C$ and endpoints $A$ and $B$ using the distance formula: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.         - AC Distance: $AC = \sqrt{(-6 - (-5))^2 + (-5 - 4)^2} = \sqrt{(-1)^2 + (-9)^2} = \sqrt{1 + 81} = \sqrt{82}$.         - BC Distance: $BC = \sqrt{(-6 - 3)^2 + (-5 - (-6))^2} = \sqrt{(-9)^2 + (1)^2} = \sqrt{81 + 1} = \sqrt{82}$.         - Observation: Because $AC = BC$, this supports the answer in part (b) based on the geometric principle that any point on the perpendicular bisector of a line segment is equidistant from the segment's endpoints.

• Parallel Lines:     - For a line to be parallel to another, it must share the same slope ($m$).     - An equation passing through $(-2, 4)$ parallel to a given line will use the point-slope formula: $y - 4 = m(x - (-2))$.

• Standard Form:     - Standard form of a linear equation is represented as $Ax + By = C$, where $A$, $B$, and $C$ are integers.

Systems of Equations and Algebraic Word Problems

• The Rice and Beans Problem (Ricardo's Shopping):     - Ricardo's total purchase parameters:         - Variables:             - $x$: pounds of rice.             - $y$: pounds of beans.         - Unit Costs: Rice costs $L\,12$ per pound; beans cost $L\,18$ per pound.         - Totals: Total weight was $103\,lb$; total expenditure was $L\,1467$.     - System of Equations:         - $x + y = 103$         - $12x + 18y = 1467$     - Solution:         - Ricardo bought $64.5\,lb$ of rice.         - Ricardo bought $38.5\,lb$ of beans.

Algebraic Simplification and Quadratic Equations

• Simplifying Expressions:     - Expressions must be simplified such that they contain no negative exponents.     - Key rule: $x^{-a} = \frac{1}{x^a}$.

• Quadratic Equations:     - Methods for solving quadratic equations ($ax^2 + bx + c = 0$) include factoring, using the quadratic formula ($x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$), or completing the square.

Financial Mathematics: Interest, Inflation, and Depreciation

• Investment and Present Value (Eva's Investment):     - Goal: Have $L\,500,000$ in 3 years ($t = 3$).     - Rate ($r$): $12\%$ or $0.12$.     - Compounding: Monthly ($n = 12$).     - Calculation involves finding the Present Value ($PV$) required to reach a Future Value ($FV$): $PV = \frac{FV}{(1 + \frac{r}{n})^{nt}}$.

• Investment Options Comparison (Pepe's Investment):     - Capital to invest: $L\,30,000$.     - Option A: 5 years at $10\%$ interest rate compounded weekly.         - Benefit/Interest generated: $L\,19,437.89$.     - Option B: 4 years at $12\%$ interest rate compounded monthly.         - Benefit/Interest generated: $L\,18,366.78$.     - Recommendation: Pepe should choose Option A as it creates a higher total interest benefit.

• Inflation and Real Value (Nyah's Promised Funds):     - Future amount: $\$100,000$ in 5 years.     - Estimated Inflation Rate ($r$): $4.5\%$ or $0.045$.     - Purpose: To find what that future money is worth in today's purchasing power (real value).     - Calculation: $Real\,Value = \frac{100000}{(1 + 0.045)^5}$.     - Result: The real value of this money today is $\$80,245.10$.

• Depreciation (Valentina's Industrial Dishwasher):     - Selling Price (Value after 5 years): $\$3,350.44$.     - Time ($t$): $5$ years.     - Annual Depreciation Rate ($r$): $11\%$ or $0.11$.     - Formula provided by context: $V = P(1 - r)^t$, where $P$ is the original price.     - $\$3,350.44 = P(1 - 0.11)^5$     - Result: Valentina originally bought the dishwasher for $\$6,000$.

Sequences and Series

• Distinguishing Sequences:     - Geometric Sequence: Identified by a common ratio ($r$). Term progression involves multiplication by a constant factor.     - Arithmetic Sequence: Identified by a common difference ($d$). Term progression involves adding a constant value.

• Case Study 1: Sequence $-3, 2, 7, \dots, 382$:     - Type: Arithmetic sequence.     - Justification: There is a common difference ($d$) of $5$ (e.g., $2 - (-3) = 5$ and $7 - 2 = 5$).     - Number of Terms ($n$):         - Formula: $a_n = a_1 + (n - 1)d$         - $382 = -3 + (n - 1)5$         - $385 = (n - 1)5$         - $77 = n - 1 \Rightarrow n = 78$.         - There are 78 terms in this sequence.     - nth Term Formula: $a_n = -3 + (n - 1)5$.     - Sum Formula ($S_n$): $S_n = \frac{n}{2}(a_1 + a_n)$.     - Total Sum: $S_{78} = \frac{78}{2}(-3 + 382) = 39 \times 379 = 14,781$.

• Case Study 2: Sequence $1/2, 1, 2, \dots, 16384$:     - Type: Geometric sequence.     - Justification: There is a common ratio ($r$) of $2$ (e.g., $1 / (1/2) = 2$ and $2 / 1 = 2$).     - Number of Terms ($n$):         - Formula: $a_n = a_1 \times r^{(n-1)}$         - $16384 = \frac{1}{2} \times 2^{(n-1)}$         - $32768 = 2^{(n-1)}$         - Since $2^{15} = 32768$, then $n - 1 = 15 \Rightarrow n = 16$.         - There are 16 terms in this sequence.     - nth Term Formula: $a_n = \frac{1}{2} \times 2^{(n-1)}$ or $a_n = 2^{n-2}$.     - Sum Formula ($S_n$): $S_n = \frac{a_1(r^n - 1)}{r - 1}$.     - Total Sum: $S_{16} = \frac{0.5(2^{16} - 1)}{2 - 1} = 0.5(65535) = 32,767.5$.

Geometry: Area of 2D Shapes and Shaded Regions

• Problem 22: Shaded area calculated as $A = 51\,in^2$ and approximately $160.22\,in^2$ (referring to two distinct components or calculations).

• Problem 23: Shaded area calculation involving a square and a circle subtraction.     - Formula: $A = (256 - 64\pi)\,cm^2$.     - Approximate Decimal Value: $A \approx 54.94\,cm^2$.

• Problem 24: Calculated Area: $A = 21\,cm^2$.

• Problem 25: Calculated Area: $A = 60\,u^2$.

Geometry: Circles, Sectors, and Surface Area of 3D Solids

• Sectors of Circles:     - Case 1 (Radius $10\,ft$, Arc $120^{\circ}$):         - Area of sector: $A = \frac{\theta}{360} \times \pi r^2 = \frac{120}{360} \times \pi (10)^2 = \frac{1}{3} \times 100\pi \approx 104.72\,ft^2$.         - Perimeter of sector: Length of arc + 2 radii.     - Case 2 (Arc Length $12\,m$, Radius $18\,m$):         - Arc length formula: $s = r\theta$, where $\theta$ is in radians.         - Angle measure in degrees: $\frac{12}{18} = \frac{2}{3}$ radians.         - Area of Sector: $A = \frac{1}{2}r^2\theta = \frac{1}{2} \times (18)^2 \times \frac{2}{3} = \frac{1}{2} \times 324 \times \frac{2}{3} = 108\,m^2$.

• Surface Area of Solids:     - Cube: With side length $s = 24\,inches$.         - Formula: $SA = 6s^2 = 6 \times (24)^2 = 6 \times 576 = 3456\,in^2$.     - Prisms:         - Prism SA (Problem 29): $115.86\,yd^2$.         - Prism SA (Problem 30): $268.8\,in^2$.     - Pyramids:         - Pyramid SA (Problem 31): $50.97\,m^2$.         - Pyramid SA (Problem 32): $47.92\,m^2$.