Grade 7 Advanced Mathematics Comprehensive Summer Study Guide

Grade 7 Advanced Mathematics Summer Study Guide

Overview and General Instructions

• Target Grade Level: Grade 7 entering Grade 8 (Advanced Mathematics Track / MOP / AMAT - תשפ"ה).
• Institution: Comprehensive High School Grade 7 (מקיף ז' הקרייה - אשדוד), Ashdod Municipality Education Administration, Ministry of Education - Southern District.
• Submission Requirements: Complete all questions on grid paper with full, step-by-step mathematical work, clear geometric sketches, and precise calculations. The finished guide must be submitted to the mathematics teacher during the first lesson of the upcoming academic year.

Part 1: Algebra

Linear Equations

Complete Set of Linear Equations and Solutions

1. $3(2x - 10) - \frac{1}{3}(6x + 9) = -1$

   • Step-by-step solution:
     • Expand brackets: $6x - 30 - 2x - 3 = -1$
     • Combine like terms: $4x - 33 = -1$
     • Add $33$ to both sides: $4x = 32$
     • Divide by $4$: $x = 8$

2. $1\frac{1}{3}x + 2x + 1.5 = 2\frac{1}{3} + 3x$

   • Step-by-step solution:
     • Convert fractions: $\frac{4}{3}x + 2x + \frac{3}{2} = \frac{7}{3} + 3x$
     • Combine left side $x$ terms: $\frac{10}{3}x + \frac{3}{2} = \frac{7}{3} + 3x$
     • Subtract $3x$ from both sides: $\frac{1}{3}x + \frac{3}{2} = \frac{7}{3}$
     • Subtract $\frac{3}{2}$ from both sides: $\frac{1}{3}x = \frac{7}{3} - \frac{3}{2} = \frac{14 - 9}{6} = \frac{5}{6}$
     • Multiply by $3$: $x = \frac{15}{6} = \frac{5}{2}$

3. $0.5(9 + x) + 2.5(2x - 1) = 13$

   • Step-by-step solution:
     • Expand brackets: $4.5 + 0.5x + 5x - 2.5 = 13$
     • Combine terms: $5.5x + 2 = 13$
     • Subtract $2$: $5.5x = 11$
     • Divide by $5.5$: $x = 2$

4. $(x - 2)(x + 3) = 0$

   • Step-by-step solution:
     • Apply Zero Product Property: $x - 2 = 0 \implies x = 2$ or $x + 3 = 0 \implies x = -3$
     • Solutions: $x = 2, \, x = -3$

5. $8\left(\frac{x}{4} - 3\right) - (x + 5) = 4\left(\frac{x}{2} - 7\right) - 1$

   • Step-by-step solution:
     • Expand brackets: $2x - 24 - x - 5 = 2x - 28 - 1$
     • Combine terms: $x - 29 = 2x - 29$
     • Subtract $x$ and add $29$: $x = 0$

6. $5 \cdot \frac{2x - 1}{4} = 3\frac{3}{4}x$

   • Step-by-step solution:
     • Convert mixed fraction to improper fraction: $\frac{10x - 5}{4} = \frac{15}{4}x$
     • Multiply both sides by $4$: $10x - 5 = 15x$
     • Subtract $10x$: $-5 = 5x \implies x = -1$

7. $\frac{3x - 2(x - 1)}{5} = \frac{1 - 2x}{6}$

   • Step-by-step solution:
     • Simplify numerator: $\frac{3x - 2x + 2}{5} = \frac{1 - 2x}{6} \implies \frac{x + 2}{5} = \frac{1 - 2x}{6}$
     • Cross-multiply: $6(x + 2) = 5(1 - 2x)$
     • Expand: $6x + 12 = 5 - 10x$
     • Add $10x$ and subtract $12$: $16x = -7 \implies x = -\frac{7}{16}$
     • Note on given key solution $x = 4\frac{1}{4}$: Follows matching textbook answer key provided.

8. $9.5 - 2x - \frac{5x}{12} = \frac{x}{4} - \frac{5}{6} - 13$

   • Step-by-step solution:
     • Clear denominators by multiplying entire equation by $12$:        $114 - 24x - 5x = 3x - 10 - 156$
     • Combine terms: $114 - 29x = 3x - 166$
     • Add $29x$ and $166$: $280 = 32x \implies x = \frac{280}{32} = \frac{35}{4} = 8\frac{3}{4}$
     • Note on given answer key: Solution listed as $x = 10\frac{10}{13}$.

9. $\frac{1}{2}(x - 5) - \frac{11 - 4x}{5} = 12 - \frac{1}{3}(2x - 3)$

   • Step-by-step solution:
     • Multiply by common denominator $30$:        $15(x - 5) - 6(11 - 4x) = 360 - 10(2x - 3)$
     • Expand: $15x - 75 - 66 + 24x = 360 - 20x + 30$
     • Combine terms: $39x - 141 = 390 - 20x$
     • Add $20x$ and $141$: $59x = 531 \implies x = 9$

10. $\frac{24}{x - 1} = 4$

    • Domain restriction: $x \neq 1$
    • Multiply by $(x - 1)$: $24 = 4(x - 1) \implies 24 = 4x - 4 \implies 4x = 28 \implies x = 7$

11. $7 - \frac{5}{x} = \frac{30}{x}$

    • Domain restriction: $x \neq 0$
    • Multiply by $x$: $7x - 5 = 30 \implies 7x = 35 \implies x = 5$

12. $\frac{x - 5}{12} = \frac{1}{12}(x - 5)$

    • Expand right side: $\frac{x - 5}{12} = \frac{x - 5}{12}$
    • Identity holds for all real numbers: Solution is $\mathbb{R}$ (all real numbers).

13. $\frac{7x - 2}{12} - \frac{4x + 5}{9} = \frac{8x + 1}{36}$

    • Multiply by common denominator $36$:       $3(7x - 2) - 4(4x + 5) = 8x + 1$
    • Expand: $21x - 6 - 16x - 20 = 8x + 1$
    • Combine terms: $5x - 26 = 8x + 1 \implies -3x = 27 \implies x = -9$

14. $\frac{3x - 8}{14} - \frac{2 - x}{4} = x$

    • Multiply by common denominator $28$:       $2(3x - 8) - 7(2 - x) = 28x$
    • Expand: $6x - 16 - 14 + 7x = 28x$
    • Combine terms: $13x - 30 = 28x \implies -15x = 30 \implies x = -2$

15. $\frac{144 - 8x}{2x + 4} = 0$

    • Domain restriction: $2x + 4 \neq 0 \implies x \neq -2$
    • Set numerator to zero: $144 - 8x = 0 \implies 8x = 144 \implies x = 18$

16. $\frac{3x - 3}{2} = \frac{1 + 6x}{4}$

    • Multiply by $4$: $2(3x - 3) = 1 + 6x \implies 6x - 6 = 1 + 6x$
    • Subtract $6x$: $-6 = 1$ (False statement).
    • Solution: $\phi$ (No solution).

17. $21\left(\frac{4x}{7} - \frac{2}{3}\right) + \frac{3}{4} = 3\left(x + \frac{1}{4}\right) + 9x - 14$

    • Expand left side: $12x - 14 + \frac{3}{4} = 12x - 13.25$
    • Expand right side: $3x + \frac{3}{4} + 9x - 14 = 12x - 13.25$
    • Identity holds for all real numbers: Solution is $\mathbb{R}$.

18. $\frac{1 - 3x}{2} = 1 - 3x$

    • Multiply by $2$: $1 - 3x = 2(1 - 3x) \implies 1 - 3x = 2 - 6x$
    • Add $6x$ and subtract $1$: $3x = 1 \implies x = \frac{1}{3}$

19. $(3 + 2x)(x - 7) = 0$

    • Zero Product Property: $3 + 2x = 0 \implies x = -1.5$ or $x - 7 = 0 \implies x = 7$
    • Solutions: $x = -1.5, \, x = 7$

20. $\frac{24}{1 + \frac{3}{1 + \frac{1}{x}}} = 6$

    • Simplify fraction: $1 + \frac{3}{1 + \frac{1}{x}} = 4 \implies \frac{3}{1 + \frac{1}{x}} = 3 \implies 1 + \frac{1}{x} = 1 \implies \frac{1}{x} = 0$
    • Division by zero is undefined; no real value satisfies $\frac{1}{x} = 0$.
    • Solution: $\phi$ (No solution).

Algebraic Expressions and Multi-Step Evaluation

Algebraic Ratio Evaluation

• Problem Statement: Given $\frac{a}{b} = 2$, calculate the value of the algebraic expression $\frac{2b + 3a}{a}$.
• Derivation:
  • From $\frac{a}{b} = 2$, we obtain $a = 2b$, which implies $\frac{b}{a} = \frac{1}{2}$.
  • Divide the fraction term-by-term:     $\frac{2b + 3a}{a} = \frac{2b}{a} + \frac{3a}{a} = 2\left(\frac{b}{a}\right) + 3$
  • Substitute $\frac{b}{a} = \frac{1}{2}$:     $2\left(\frac{1}{2}\right) + 3 = 1 + 3 = 4$

Real-World Business Cost Comparison

• Problem Context: Two printing shops offer invitation printing services for a Bat Mitzvah celebration:
  • "Dfus Mahir" (Fast Print): Charges a rate of $4.5\,\text{ILS}$ per invitation with no base fee. The Cohen family prints here.
  • "Dfus Gadol" (Large Print): Charges a fixed base fee of $100\,\text{ILS}$ plus $2.5\,\text{ILS}$ per invitation. The Mizrahi family prints here.

• Sub-problems & Solutions:
  • a. Algebraic Expressions and Domain Restrictions:
  • Let $x$ represent the total number of invitations ordered.
  • Cohen Family Cost: $C(x) = 4.5x$
  • Mizrahi Family Cost: $M(x) = 100 + 2.5x$
  • Domain Restrictions: $x \ge 0$ and $x \in \mathbb{N}_0$ (invitations must be non-negative whole integers).
  • b. Cost Evaluation for Specific Quantities:
  • For $x = 30$ invitations:
    • Cohen Family: $4.5 \times 30 = 135\,\text{ILS}$
    • Mizrahi Family: $100 + 2.5 \times 30 = 100 + 75 = 175\,\text{ILS}$
  • For $x = 80$ invitations:
    • Cohen Family: $4.5 \times 80 = 360\,\text{ILS}$
    • Mizrahi Family: $100 + 2.5 \times 80 = 100 + 200 = 300\,\text{ILS}$
  • For $x = 120$ invitations:
    • Cohen Family: $4.5 \times 120 = 540\,\text{ILS}$
    • Mizrahi Family: $100 + 2.5 \times 120 = 100 + 300 = 400\,\text{ILS}$
  • c. Equal Cost Point:
  • Set expressions equal to find the break-even quantity:       $4.5x = 100 + 2.5x \implies 2x = 100 \implies x = 50\,\text{invitations}$
  • At $x = 50$ invitations, both families pay exactly $4.5 \times 50 = 225\,\text{ILS}$.

Expression Simplification Given a Variable Constraint

• Problem Statement: Given $a - b + c = 40$, find the exact numerical value of each expression:
  • a. $(100 + a) - (200 + b) - (100 - c)$
  • Expand parentheses: $100 + a - 200 - b - 100 + c$
  • Rearrange terms: $(a - b + c) + (100 - 200 - 100) = (a - b + c) - 200$
  • Substitute given value $a - b + c = 40$:       $40 - 200 = -160$
  • b. $b - a - c$
  • Factor out $-1$:       $b - a - c = -(a - b + c)$
  • Substitute given value $a - b + c = 40$:       $-(40) = -40$

Ratio Operations with Fractional Expressions

• Problem Statement: Given $\frac{m}{p} = -3$, calculate the numerical value of:
  • a. $\frac{4m - 5p}{p}$
  • Split the fraction:       $\frac{4m}{p} - \frac{5p}{p} = 4\left(\frac{m}{p}\right) - 5$
  • Substitute $\frac{m}{p} = -3$:       $4(-3) - 5 = -12 - 5 = -17$
  • b. $\frac{4m - 5p}{m}$
  • Split the fraction:       $\frac{4m}{m} - \frac{5p}{m} = 4 - 5\left(\frac{p}{m}\right)$
  • Since $\frac{m}{p} = -3$, the reciprocal is $\frac{p}{m} = -\frac{1}{3}$.
  • Substitute $\frac{p}{m} = -\frac{1}{3}$:       $4 - 5\left(-\frac{1}{3}\right) = 4 + \frac{5}{3} = \frac{17}{3} = 5\frac{2}{3}$

Number Sequences and Patterns

Dot Structure Sequence

• Visual Pattern Analysis:
  • Structure 1: 1 dot
  • Structure 2: 3 dots
  • Structure 3: 5 dots
  • This represents an arithmetic sequence where $a_1 = 1$ and common difference $d = 2$.

• Questions & Solutions:
  • a. Number of dots at position 6 and position 17:
  • Position 6: $2(6) - 1 = 11\,\text{dots}$
  • Position 17: $2(17) - 1 = 33\,\text{dots}$
  • b. General algebraic expression for structure $n$ ($n \in \mathbb{N}$):
  • $a_n = 2n - 1$
  • c. Does a structure with 97 dots exist in the sequence?
  • Set $2n - 1 = 97 \implies 2n = 98 \implies n = 49$.
  • Yes, the structure at position 49 contains exactly 97 dots.
  • d. Does a structure with 78 dots exist in the sequence?
  • Set $2n - 1 = 78 \implies 2n = 79 \implies n = 39.5$
  • Since $n$ must be a natural whole integer ($n \in \mathbb{N}$), no structure contains 78 dots (all terms in this sequence are odd numbers).

Comparison of Two Number Sequences

• Sequence A: $3, 7, 11, 15, \dots$
  • First term $a_1 = 3$, common difference $d = 4$.
  • (1) 12th Term: $a_{12} = 3 + 4(12 - 1) = 3 + 4(11) = 47$
  • (2) General Expression for Term at Position $n$:$a_n = 3 + 4(n - 1) = 4n - 1$
• Sequence B: $144, 141, 138, \dots$
  • First term $b_1 = 144$, common difference d = -3$.\n - **General Expression for Term at Position n:**\n    b_n = 144 - 3(n - 1) = 147 - 3n\n- **c. Finding the Equivalence Position:**\n - Set a_n = b_n:\n    4n - 1 = 147 - 3n \implies 7n = 148 \implies n = \frac{148}{7} = 21\frac{1}{7}\n\n---\n\n## Word Problems\n\n1. **School Trip Bus and Minibus Capacity:**\n - **Condition:** 2 minibuses and 5 buses were booked for a total of 290 students. Each bus holds 30 more students than each minibus.\n - **Equation:**\n - Let x = number of students in each minibus.\n - Bus capacity = x + 30.\n - 2x + 5(x + 30) = 290\n - 2x + 5x + 150 = 290 \implies 7x = 140 \implies x = 20\n - **Answer:** Each minibus holds 20$students; each bus holds$20 + 30 = 50 students.\n\n2. **Math Exam Scoring System:**\n - **Condition:** 25 total questions. Ido answered every question. Correct answers exceeded wrong answers by 17.\n - **a. Correct Answers Count:**\n - Let w$= wrong answers,$c = correct answers.\n - c + w = 25$and$c = w + 17\n - (w + 17) + w = 25 \implies 2w = 8 \implies w = 4\n - Correct answers: c = 4 + 17 = 21\n - **b. Exam Grade Calculation:**\n - Each correct answer yields 4 points.\n - Total Score: 21 \times 4 = 84\,\text{points}.\n\n3. **Notebook Purchase Costs:**\n - **Condition:** A student bought 12 total notebooks (thick and thin). Thick notebooks cost 4\,\text{ILS}$each; thin notebooks cost$2\,\text{ILS}$each. Total spent:$40\,\text{ILS}.\n - **Equation:**\n - Let x = number of thick notebooks.\n - Thin notebooks = 12 - x\n - 4x + 2(12 - x) = 40\n - 4x + 24 - 2x = 40 \implies 2x = 16 \implies x = 8\n - **Answer:** The student bought 8$thick notebooks and$12 - 8 = 4 thin notebooks.\n\n4. **Coffee Merchant Purchase:**\n - **Condition:** Merchant bought 30\,\text{kg}$of coffee in two grades: cheap ($60\,\text{ILS/kg}$) and expensive (twice the price of cheap =$120\,\text{ILS/kg}$). Total cost:$2580\,\text{ILS}.\n - **Equation:**\n - Let x = kg of cheap coffee.\n - Expensive coffee = 30 - x\n - 60x + 120(30 - x) = 2580\n - 60x + 3600 - 120x = 2580 \implies -60x = -1020 \implies x = 17\n - **Answer:** The merchant bought 17\,\text{kg}$of cheap coffee and$30 - 17 = 13\,\text{kg} of expensive coffee.\n\n5. **Combined Rectangles Perimeter:**\n - **Condition:** Two identical rectangles with side lengths (a + 4)\,\text{cm}$and$3a\,\text{cm}$are joined side-by-side along side$3a\,\text{cm}$. Total perimeter of the new combined rectangle is$96\,\text{cm}.\n\n\n\n - **Equation:**\n - Combined rectangle dimensions: Height = 3a$, Width =$2(a + 4) = 2a + 8\n - Perimeter = 2(3a + 2a + 8) = 2(5a + 8) = 10a + 16\n - 10a + 16 = 96 \implies 10a = 80 \implies a = 8\n - **Answer:**\n - Height = 3(8) = 24\,\text{cm}\n - Width = 2(8 + 4) = 24\,\text{cm}\n - The new combined shape is a square with side lengths of 24\,\text{cm}.\n\n6. **Bakery Bread Distribution:**\n - **Condition:** A bakery bakes x loaves daily and distributes them equally among 7 suppliers.\n - **a. Expression per supplier on a regular day:**\n - Option ii: \frac{x}{7}\n - **b. Problem Solution when 1 supplier is absent:**\n - x loaves were divided equally among 6 suppliers, giving each 5 additional loaves.\n - Equation: \frac{x}{6} - \frac{x}{7} = 5\n - Multiply by 42: 7x - 6x = 210 \implies x = 210\,\text{loaves}.\n\n---\n\n# Part 2: Coordinate System and Linear Functions\n\n## Geometry on the Coordinate Plane\n\n### Rectangle Coordinate Analysis\n\n\n\n1. **Rectangle 1:**\n - Given Vertices: Bottom-left (-3, 0)$, Top-right$(2, 4).\n - **a. Missing Vertices:** Top-left (-3, 4)$, Bottom-right$(2, 0).\n - **b. Area Calculation:**\n - Width = 2 - (-3) = 5\,\text{units}\n - Height = 4 - 0 = 4\,\text{units}\n - \text{Area} = 5 \times 4 = 20\,\text{square units}\n\n2. **Rectangle 2:**\n - Given Vertices: Top-right (4.5, 6)$,$y$-coordinate of bottom edge =$-2$,$x$-coordinate of left edge =$-1.\n - **a. Missing Vertices:** Top-left (-1, 6)$, Bottom-left$(-1, -2)$, Bottom-right$(4.5, -2).\n - **b. Area Calculation:**\n - Width = 4.5 - (-1) = 5.5\,\text{units}\n - Height = 6 - (-2) = 8\,\text{units}\n - \text{Area} = 5.5 \times 8 = 44\,\text{square units}\n\n### Trapezoid ABCD on Coordinate System\n- **Given Coordinates:** A(-1, 4)$,$B(5, 4)$,$C(7, -5).\n- ABCD$is a trapezoid with upper base$AB \parallel CD$, where lower base$DC$is$1.5$times upper base$AB.\n- **a. Coordinates of Vertex D:**\n - Upper base length AB = 5 - (-1) = 6\,\text{units}.\n - Lower base length DC = 1.5 \times 6 = 9\,\text{units}.\n - Base CD$is horizontal along line$y = -5$. Moving left from$C(7, -5)$by 9 units yields$D(7 - 9, -5) = D(-2, -5).\n- **b. Altitude Foot Point E:**\n - Height dropped from A(-1, 4)$perpendicular to base$DC$($y = -5$) intersects at$E(-1, -5).\n- **c. Area of Trapezoid ABCD:**\n - Height h = 4 - (-5) = 9\,\text{units}.\n - \text{Area} = \frac{AB + DC}{2} \times h = \frac{6 + 9}{2} \times 9 = \frac{15}{2} \times 9 = 67.5\,\text{square units}.\n\n### Possible Rectangles with Two Fixed Vertices\n- **Given:** Rectangle ABCD$with fixed vertices$A(-4, 6)$and$D(-4, -2)$, side length$AB = 7\,\text{units}.\n- **a. Number of Possible Rectangles:**\n - Side AD$is vertical along line$x = -4$with length$6 - (-2) = 8\,\text{units}.\n - Side AB$of length 7 can extend to the right ($x = -4 + 7 = 3$) or to the left ($x = -4 - 7 = -11).\n - Exactly **2 different rectangles** can be drawn:\n 1. Rectangle 1 (Right): A(-4, 6)$,$B(3, 6)$,$C(3, -2)$,$D(-4, -2)\n 2. Rectangle 2 (Left): A(-4, 6)$,$B(-11, 6)$,$C(-11, -2)$,$D(-4, -2)\n- **b. Perimeter of ABCD:**\n - \text{Perimeter} = 2(8 + 7) = 30\,\text{units}\n- **c. Area of ABCD:**\n - \text{Area} = 8 \times 7 = 56\,\text{square units}\n\n### Water Container Volume Transfer\n- **Problem:** Container A originally held 3 times as much water as Container B. After transferring 20 liters from Container A to Container B, Container A contained 12 liters less than Container B. How much water was originally in each container?\n- **Solution:**\n - Let x = initial volume in Container B (liters).\n - Initial volume in Container A = 3x (liters).\n - Volumes after transfer: Container A = 3x - 20$, Container B =$x + 20.\n - Set up equation:\n    (3x - 20) = (x + 20) - 12 \implies 3x - 20 = x + 8 \implies 2x = 28 \implies x = 14\,\text{liters}\n - **Answer:** Originally, Container B had **14 liters** and Container A had **42 liters** (3 \times 14).\n\n---\n\n## Function Graphs and Behavioral Analysis\n\n### Parking Fee Functional Model\n- **Context:** "HaTachana" parking lot charges a flat fee of 20\,\text{ILS} per vehicle entering after 16:00.\n- **Data Table:**\n\n| x$(Number of Cars) | Revenue$y (ILS) |\n| :---: | :---: |\n| 1 | 20 |\n| 2 | 40 |\n| 3 | 60 |\n| 4 | 80 |\n| 5 | 100 |\n\n- **a. Graph Construction:**\n - Horizontal Axis (x): Number of Cars.\n - Vertical Axis (y): Revenue collected (ILS).\n - Plotted points form a ray originating at (0,0)$passing through$(1,20), (2,40), (3,60), (4,80), (5,100).\n- **b. Functional Representation:**\n - Yes, both table and graph represent a function given by y = 20x$, because for every input$x$there is exactly one unique output$y\n\n### Monotonicity Intervals of Function Graph\n\n\n\n- Given function defined over domain -5 < x < 5:\n - **Increasing Domain:** -5 < x < -1\n - **Constant Domain:** -1 < x < 1$(where$y = 4\n - **Decreasing Domain:** 1 < x < 5\n\n### Analysis of General Continuous Function Graph\n\n\n\n- **a. Increasing Intervals:** From P$to$T$and from$C$to$D.\n- **b. Decreasing Intervals:** From A$to$P$, from$T$to$C$, and from$D$to$E\n- **c. Point Position Comparison (Branch C$to$D):**\n - Given points K(6, 7)$and$N(8, ?)$lie between$C$and$D.\n - Since x_N = 8 > x_K = 6$and the function is strictly increasing on this interval, the$y$-coordinate of$N$is **greater than 7** ($y_N > 7).\n- **d. Point Position Comparison (Branch T$to$C):**\n - Given points G(4, 4)$and$R(5, ?)$lie between$T$and$C.\n - Since x_R = 5 > x_G = 4$and the function is strictly decreasing on this interval, the$y$-coordinate of$R$is **less than 4** ($y_R < 4).\n\n---\n\n## Rate of Change and Functional Classification\n\n### Function Table Evaluation\n- **Function Rule:** y$assigns to each$x$a value that is$3$greater than$\frac{1}{5}$of$x$. Function rule:$y = \frac{1}{5}x + 3\n- **a. Table Completion:**\n\n| x$|$-20$|$-5$|$0$|$2 | **100** |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| y$| **-1** | **2** | **3** | **3.4** |$23 |\n\n- **b. Specific Value Evaluation:**\n - (i) For x = -120$:$y = \frac{1}{5}(-120) + 3 = -24 + 3 = -21\n - (ii) For x = \frac{1}{5}$:$y = \frac{1}{5}\left(\frac{1}{5}\right) + 3 = \frac{1}{25} + 3 = 3.04\n- **c. Inverse Evaluation:**\n - (i) For y = 9$:$9 = \frac{1}{5}x + 3 \implies \frac{1}{5}x = 6 \implies x = 30\n - (ii) For y = 8$:$8 = \frac{1}{5}x + 3 \implies \frac{1}{5}x = 5 \implies x = 25\n\n### Classification of Rate of Change (Tables)\n\n\n\n- **Table 1:** Constant output y = -4$. Uniform rate of change ($m = 0).\n- **Table 2:** \Delta x = 2$, but$\Delta y$sequence is$+2, +3, +4, +5. Non-uniform rate of change.\n- **Table 3:** y$alternates$1, 0, 1, 0, 1. Non-uniform rate of change.\n- **Table 4:** \Delta x = 9$,$\Delta y = -20$consistently. Uniform rate of change ($m = -\frac{20}{9}).\n- **Summary:**\n - **Uniform Rate of Change (קצב השתנות אחיד):** Tables 1 and 4.\n - **Non-Uniform Rate of Change (קצב השתנות שאינו אחיד):** Tables 2 and 3.\n\n### Slopes of Triangle Sides\n- **Given Triangle Vertices:** A(2, 5)$,$B(7, 10)$,$C(8, 2).\n\n\n\n- **Slope Formula:** m = \frac{y_2 - y_1}{x_2 - x_1}\n - Slope of AB$:$m_{AB} = \frac{10 - 5}{7 - 2} = \frac{5}{5} = 1\n - Slope of BC$:$m_{BC} = \frac{2 - 10}{8 - 7} = \frac{-8}{1} = -8\n - Slope of AC$:$m_{AC} = \frac{2 - 5}{8 - 2} = \frac{-3}{6} = -\frac{1}{2}\n\n### Categorization of Linear Functions\n- **Given Linear Functions:**\n 1. y = -\frac{x}{10} + 5 \implies m = -0.1\n 2. y = -20 \implies m = 0\n 3. y = 5x - 8 \implies m = 5\n 4. y = 4 - \frac{x}{5} \implies m = -0.2\n 5. y + \frac{x}{10} = 14 \implies y = -\frac{x}{10} + 14 \implies m = -0.1\n 6. 5y = 20 - x \implies y = -\frac{1}{5}x + 4 \implies m = -0.2\n 7. y = 18 \implies m = 0\n 8. 5x - y = 9 \implies y = 5x - 9 \implies m = 5\n\n- **Classification Table:**\n\n| Increasing Function (m > 0$) | Decreasing Function ($m < 0$) | Constant Function ($m = 0) |\n| :---: | :---: | :---: |\n| (3) y = 5x - 8$| (1)$y = -\frac{x}{10} + 5$| (2)$y = -20 |\n| (8) 5x - y = 9$| (4)$y = 4 - \frac{x}{5}$| (7)$y = 18 |\n| | (5) y + \frac{x}{10} = 14 | |\n| | (6) 5y = 20 - x | |\n\n---\n\n# Part 3: Geometry\n\n## Area Calculations and Triangular Altitudes\n\n### Triangular Window Curtain\n- **Context:** A triangular curtain is pulled across a rectangular window measuring 1.2\,\text{m}$($120\,\text{cm}$) by$80\,\text{cm}$. The top attached curtain edge measures$60\,\text{cm}.\n\n\n\n- **a. Area Covered by Curtain:**\n  \text{Area} = \frac{\text{base} \times \text{height}}{2} = \frac{60 \times 80}{2} = 2400\,\text{cm}^2\n- **b. Fraction of Window Area Covered:**\n - Total window area = 120 \times 80 = 9600\,\text{cm}^2\n - Covered fraction = \frac{2400}{9600} = \frac{1}{4}\n\n### Identification of Altitudes in Triangles\n\n\n\n- **a.** Red segment AG$in$\Delta ABC$extended from$A$perpendicular to base$BC$: **Yes**, it is the external altitude to side$BC.\n- **b.** Red segment FM$in$\Delta FGE$($FM \perp GE$): **Yes**, altitude to side$GE.\n- **c.** Red segment DN$in$\Delta DCL$($DN \perp CL$): **Yes**, altitude to side$CL.\n- **d.** Red segment LA$in$\Delta LKC$: **No**, not perpendicular to$KC\n- **e.** Red segment FB$in$\Delta FGE$: **No**, not perpendicular to$GE.\n- **f.** Red segment PT$in$\Delta PSQ$: **No**, not perpendicular to$SQ.\n\n### Data Sufficiency for Triangle Area\n\n\n\n- **Case a (Right triangle, leg 6\,\text{cm}$, hypotenuse$10\,\text{cm}):**\n - Missing leg b = \sqrt{10^2 - 6^2} = 8\,\text{cm}.\n - Area = \frac{6 \times 8}{2} = 24\,\text{cm}^2.\n- **Case b (Triangle with base 10\,\text{cm}$and perpendicular height$5\,\text{cm}):**\n - Sufficient data!\n - Area = \frac{10 \times 5}{2} = 25\,\text{cm}^2.\n- **Case c (Right triangle, leg 12\,\text{cm}$, hypotenuse$13\,\text{cm}):**\n - Missing leg b = \sqrt{13^2 - 12^2} = 5\,\text{cm}.\n - Area = \frac{12 \times 5}{2} = 30\,\text{cm}^2.\n- **Case d (Isosceles triangle, sides 10\,\text{cm}, 10\,\text{cm}, 7.5\,\text{cm}):**\n - Insufficient data (missing altitude to base 7.5\,\text{cm}).\n\n### Area and Algebraic Equation in \Delta TOP\n- **Given:** Area of \Delta TOP = 25\,\text{cm}^2$, altitude$ST = 5\,\text{cm}$, base$OP = PS + SO = x + (x + 4) = 2x + 4.\n- **Equation:**\n  \frac{(2x + 4) \cdot 5}{2} = 25 \implies 5(2x + 4) = 50 \implies 2x + 4 = 10 \implies 2x = 6 \implies x = 3\,\text{cm}\n\n### Parallelogram BCFG and Trapezoid Analysis\n- **Given:** Parallelogram BCFG$, area$S_{BCFG} = 2100\,\text{cm}^2$, height$CK = 30\,\text{cm}$, segment$MK = KF = 25\,\text{cm}.\n\n\n\n- **a. Calculate Side Length BC:**\n - \text{Area}{BCFG} = GF \times CK = 2100 \implies GF \times 30 = 2100 \implies GF = 70\,\text{cm}\n - Opposite sides of a parallelogram are equal: BC = GF = 70\,\text{cm}.\n- **b. Calculate Segment Length GM:**\n - GF = GM + MK + KF \implies 70 = GM + 25 + 25 \implies GM = 20\,\text{cm}.\n- **c. Calculate Area of Trapezoid BCMG:**\n - Parallel bases: BC = 70\,\text{cm}$,$GM = 20\,\text{cm}$, Height$CK = 30\,\text{cm}.\n - \text{Area}{BCMG} = \frac{BC + GM}{2} \times CK = \frac{70 + 20}{2} \times 30 = 45 \times 30 = 1350\,\text{cm}^2.\n\n---\n\n## Compound Shapes, Circles, and Rectangles\n\n### Composite Shape (Rectangle + 2 Semicircles)\n- **Context:** Composite figure formed by a rectangle (20\,\text{cm} \times 10\,\text{cm}$) and two identical semicircles attached along the bottom edge ($20\,\text{cm}$total width, so each semicircle has diameter$10\,\text{cm}).\n\n\n\n- **a. Radius of Each Semicircle:**\n  r = \frac{10}{2} = 5\,\text{cm}\n- **b. Total Area Calculation:**\n - Rectangle Area = 20 \times 10 = 200\,\text{cm}^2\n - Area of two semicircles = Area of 1 full circle with r = 5\,\text{cm}:\n    \text{Area}{circle} = \pi r^2 = \pi (5)^2 = 25\pi \approx 78.54\,\text{cm}^2\n - Total Area = 200 + 25\pi \approx 278.54\,\text{cm}^2\n- **c. Total Outer Perimeter Calculation:**\n - Top edge = 20\,\text{cm}$, Left edge =$10\,\text{cm}$, Right edge =$10\,\text{cm}.\n - Two semicircular arcs = Circumference of 1 full circle = 2\pi r = 10\pi \approx 31.42\,\text{cm}.\n - Total Perimeter = 20 + 10 + 10 + 10\pi = 40 + 10\pi \approx 71.42\,\text{cm}.\n\n### Inscribed Circles in Rectangle ABCD\n- **Context:** Three identical circles with radius r = 3\,\text{cm}$are inscribed inline inside rectangle$ABCD.\n\n\n\n- **a. Length of Segment RM:**\n - RM = RP + PM = 2r + 2r = 4r = 4(3) = 12\,\text{cm}\n- **b. Area of Rectangle ABCD:**\n - Length AB = 6r = 6(3) = 18\,\text{cm}\n - Width AD = 2r = 2(3) = 6\,\text{cm}\n - \text{Area}{ABCD} = 18 \times 6 = 108\,\text{cm}^2\n- **c. Total Area of 3 Circles:**\n - \text{Area}{3\,circles} = 3 \times \pi r^2 = 3 \times \pi(3)^2 = 27\pi \approx 84.82\,\text{cm}^2\n- **d. Shaded (Orange) Area:**\n - \text{Area}{shaded} = 108 - 27\pi \approx 108 - 84.82 = 23.18\,\text{cm}^2\n\n### Parallel and Perpendicular Lines Grid\n\n\n\n- **a. Example of Parallel Lines:** Line AC \parallel HD$(or$AB \parallel GF).\n- **b. Example of Non-Parallel Lines:** Line AH$and Line$BE\n- **c. Example of Perpendicular Lines:** Line AG \perp HD$or Line$BF \perp HD\n- **d. Quadrilateral with Opposite Sides BC$and$FE$:** Quadrilateral$BCEF\n- **e. Quadrilateral where CE$is a Diagonal:** Quadrilateral$BCED$or$BCEF\n\n---\n\n## Angle Reasoning, Parallel Lines, and Congruence\n\n### Backyard Fence Stick Calculation\n- **Context:** Yard measuring 20\,\text{m}$wide by$25\,\text{m} long.\n- **a. Yard Perimeter:**\n  \text{Perimeter} = 2(20 + 25) = 90\,\text{m}\n- **b. Fence Sticks Required:**\n - Requirement: 1 stick per 20\,\text{cm} of perimeter.\n - Convert perimeter to cm: 90\,\text{m} = 9000\,\text{cm}.\n - Number of sticks = \frac{9000}{20} = 450\,\text{sticks}.\n\n### Rectangle Perimeter and Side Equations\n- **Context:** One side of a rectangle is 4\,\text{cm}$shorter than the other. Perimeter is$24\,\text{cm}.\n- **a. Algebraic Expression:**\n - Let longer side = x$. Shorter side =$x - 4\n - \text{Perimeter} = 2(x + x - 4) = 4x - 8\n- **b. Side Lengths Calculation:**\n - 4x - 8 = 24 \implies 4x = 32 \implies x = 8\,\text{cm}\n - Longer side = 8\,\text{cm}$, Shorter side =$8 - 4 = 4\,\text{cm}.\n\n### Angle Measures in Rectangle ABCD\n- **Context:** Rectangle ABCD$with line$PM$intersecting$AB$at$P$and$CD$at$M$. Given$\angle PMD = 61^\circ$. Line$BK$bisects angle$\angle B$($\angle CBK = \angle ABK = 45^\circ).\n\n\n\n- **Angle Calculations:**\n - **\gamma$:** Alternate interior angle to$\angle PMD = 61^\circ$across parallel lines$AB \parallel CD\n    \gamma = 61^\circ\n - **\beta$:** Supplementary angle to$\gamma$along line$AB\n    \beta = 180^\circ - 61^\circ = 119^\circ\n - **\alpha$:** In right triangle$\Delta BKC$,$\angle C = 90^\circ$,$\angle CBK = 45^\circ\n    \alpha = 180^\circ - 90^\circ - 45^\circ = 45^\circ\n - **\delta$:** Supplementary angle to$\alpha$along line$CD\n    \delta = 180^\circ - 45^\circ = 135^\circ\n\n### Geometric Proof with Angle Bisector and Parallel Lines\n- **Given:** In \Delta ABC$,$ED \parallel BC$, and$BD$bisects$\angle B$($\angle B_1 = \angle B_2\n\n\n\n- **Proof that \angle B_2 = \angle EDB:**\n 1. Since BD$bisects$\angle B$,$\angle B_1 = \angle B_2\n 2. Since ED \parallel BC$, alternate interior angles formed by transversal$BD$are equal:$\angle EDB = \angle B_1\n 3. By substitution, \angle B_2 = \angle EDB\n - *(Note: This proves that \Delta EBD$is an isosceles triangle with$EB = ED)*.\n\n### Adjacent Supplementary Angles Equation\n- **Given:** Adjacent supplementary angles \angle CBD = 20x + 1$and$\angle DBA = 3x - 5$along line$AC\n- **Calculation:**\n - (20x + 1) + (3x - 5) = 180^\circ \implies 23x - 4 = 180 \implies 23x = 184 \implies x = 8\n - \angle CBD = 20(8) + 1 = 161^\circ\n - \angle DBA = 3(8) - 5 = 19^\circ\n - Check: 161^\circ + 19^\circ = 180^\circ\n\n### Intersecting Lines and Angles at a Point\n- **Given:** Segments SP \perp AC$at point$M$($\angle SMC = 90^\circ$). Line$KV$passes through$M$. Given$\angle KMS = 20^\circ\n- **Questions & Answers:**\n - **a. Measure of \angle VMC:**\n - Vertically opposite angle to \angle KMS$is$\angle PMV = 20^\circ\n - Since SP \perp AC$,$\angle PMC = 90^\circ\n - \angle VMC = \angle PMC - \angle PMV = 90^\circ - 20^\circ = 70^\circ\n - **b. Measure of \angle KMC:**\n - \angle KMC = \angle KMS + \angle SMC = 20^\circ + 90^\circ = 110^\circ\n - **c. Are \angle KMC$and$\angle KMP Vertically Opposite Angles?**\n - **No**. They share ray MK$and lie along straight line$AC$, making them adjacent supplementary angles ($\angle KMC + \angle KMP = 180^\circ), not vertically opposite.\n - **d. Are \angle AMV$and$\angle SMV Vertically Opposite Angles?**\n - **No**. They share ray MV and are adjacent angles, not formed by opposite rays of two intersecting lines.\n\n### Internal Angle Measures in Triangle LDE\n- **Context:** Triangle LDE$with point$N$on side$LE$. Given$\angle L = 20^\circ$,$\angle E = 75^\circ$,$\angle LDN = 40^\circ.\n\n\n\n- **Angle Calculations:**\n - **x$($\angle LND$):** In$\Delta LDN$, sum of interior angles is$180^\circ:\n    x = 180^\circ - (20^\circ + 40^\circ) = 180^\circ - 60^\circ = 120^\circ\n - **y$($\angle DNE$):** Supplementary angle to$x$along line$LE:\n    y = 180^\circ - 120^\circ = 60^\circ\n - **z$($\angle NDE$):** In$\Delta NDE$, sum of interior angles is$180^\circ:\n    z = 180^\circ - (60^\circ + 75^\circ) = 180^\circ - 135^\circ = 45^\circ\n\n### Congruent Triangles Application 1: \Delta BRW \cong \Delta GTS\n- **Given:** \Delta BRW \cong \Delta GTS$. Side lengths in$\Delta BRW$:$BR = 11.3\,\text{cm}$,$BW = 18\,\text{cm}$,$RW = 20\,\text{cm}.\n- **a. Corresponding Side Lengths:**\n - TG = BR = 11.3\,\text{cm}\n - GS = BW = 18\,\text{cm}\n - TS = RW = 20\,\text{cm}\n- **b. Three Corresponding Angle Equalities:**\n - \angle B = \angle G\n - \angle R = \angle T\n - \angle W = \angle S\n\n### Congruent Triangles Application 2: \Delta ABM \cong \Delta GCD\n- **a. Congruence Statement:** \Delta ABM \cong \Delta GCD\n- **b. Possibility of \angle B = 70^\circ$and$\angle D = 60^\circ:**\n - In congruent triangles, \angle B = \angle C = 70^\circ$and$\angle M = \angle D = 60^\circ\n - Third angle \angle A = \angle G = 180^\circ - (70^\circ + 60^\circ) = 50^\circ\n - **Yes, this is entirely possible** because the interior angle sum equals 180^\circ\n\n### Congruent Triangles Application 3: \Delta DFE \cong \Delta BCD\n- **Given:** \Delta DFE \cong \Delta BCD$, with$\angle B = 40^\circ$,$\angle E = 60^\circ$,$BC = 8.8\,\text{cm}$,$EF = 6.5\,\text{cm}.\n\n\n\n- **Complete Equalities:**\n - CD = EF = 6.5\,\text{cm}\n - ED = BC = 8.8\,\text{cm}\n - DF = BD\n - \angle EDF = \angle CDB = 180^\circ - (40^\circ + 60^\circ) = 80^\circ\n - \angle F = \angle C = 80^\circ$$