Math Exam 1

Algebraic expressions and evaluations

1. Writing or solving algebraic problems

  • Sum: x+y+23x + y + 23
  • Difference: z15z – 15
  • Combine like terms with multiplication shown: 2+r+r+s+2+t+r+t+s+t+t+t=4+3r+2s+4t2 + r + r + s + 2 + t + r + t + s + t + t + t = 4 + 3\cdot r + 2\cdot s + 4\cdot t
  • Product with given values: mn3p=74312=42m\cdot n \cdot 3p = 7\cdot 4 \cdot 3\cdot \tfrac{1}{2} = 42
  • Quotient: xy\dfrac{x}{y}
  • Power: 158
  • 2. Order of operations
  • Evaluate:843[5+2×64]+3×5284 – 3[5 + 2 \times 6 – 4] + 3 \times 5^2
    1. Inside brackets: 5+2×64=5+124=135 + 2 \times 6 – 4 = 5 + 12 – 4 = 13
    2. Multiply: 3×13=39-3 \times 13 = -39
    3. Power and multiply: 52=255^2 = 25, then 3×25=753 \times 25 = 75
    4. Combine: 8439+75=12084 – 39 + 75 = 120
  • Result: 120
  • 3. Distributive property
  • Expand: 3x(2xy+ab)=6x23xy+3abx
  • 4. Opposite and absolute value
  • For 1515: Opposite: 15-15; Absolute value: 15=15|15| = 15
  • For 9-9: Opposite: 99; Absolute value: 9=9
  • 5. Evaluate the expressions
  • Sum: 9+(4)=59 + (-4) = 5
  • Subtract a negative: 12(6)=1812 – (-6) = 18
  • Subtract: 154=19-15 – 4 = -19
  • Multiply negatives: 2×(8)=16-2 \times (-8) = 16
  • Divide: 16÷4=4
  • 6. Factor out common factors
  • Expression: 6x3+12x2y=6x2(x+2y)
  • Graphing guidance
  • 7a. Parabola y=(x+2)21y = (x + 2)^2 – 1
  • Vertex: (2,1)(-2, -1)
  • Axis of symmetry: x=2x = -2
  • Opens: Upward
  • Intercepts:
    • x-intercepts: Solve (x+2)2=1x=3,1(x+2)^2 = 1 \Rightarrow x = -3, -1
    • y-intercept: x=0y=3x=0 \Rightarrow y=3
  • 7b. Absolute value y=x2y = |x – 2|
  • Vertex: (2,0)(2, 0)
  • Shape: V with slopes +1+1 (right) and 1-1 (left)
  • Intercepts:
    • x-intercept: x=2x = 2
    • y-intercept: x=0y=2x=0 \Rightarrow y=2
  • 7c. Two lines on one graph
  • First line: y=3x+4y = 3x + 4 (slope 33, y-intercept 44)
  • Second line: The expression appears unclear. Can you confirm whether it is y=32x1y = \tfrac{3}{2}x – 1 or y=32xy = \tfrac{3}{2} – x?
  • Identifying function types
  • 8. Function types and properties
  • y=12x+5y = 12x + 5: Linear, not proportional (nonzero intercept).
  • y=4xy = 4x: Linear and proportional (direct variation with constant 44).
  • y=x3y = x^3: Cubic, odd function, passes through origin.
  • Comparing rational numbers
  • 9. Insert the correct inequality
  • Compare: 0.6670.667 \, and 85\,\dfrac{8}{5}0.667<850.667 < \dfrac{8}{5}
  • Compare: 5-5 \, and 3\,-35<3-5 < -3
  • Compare: 0.0010.001 \, and 0.01\,0.010.001<0.01
  • Rational numbers and rounding
  • 10. What is a rational number?
  • Definition: A rational number is any number that can be written as pq\dfrac{p}{q} where p,qp, q are integers and q0q \neq 0. Its decimal form terminates or repeats.
  • 11. Approximations
  • Nearest ten-thousandth: 4.5602464.56024.560246 \to 4.5602
  • Nearest hundredth: 3.141592653.143.14159265 \to 3.14
  • Nearest tenth: 32=1.51.5
  • Linear modeling and age functions
  • 13. Snow sculpture problem
  • a. Weight after xx days: W(x)=2503xW(x) = 250 – 3x
  • b. Specific times:
    • 7 days: W(7)=25021=229W(7) = 250 – 21 = 229
    • 31 days: W(31)=25093=157W(31) = 250 – 93 = 157
  • c. Solve for time:
    • 160 lb: 2503x=1603x=90x=30250 – 3x = 160 \Rightarrow 3x = 90 \Rightarrow x = 30
    • 150 lb: 2503x=1503x=100x=100333.3250 – 3x = 150 \Rightarrow 3x = 100 \Rightarrow x = \dfrac{100}{3} \approx 33.\overline{3}
  • d. Assumption: The weight decreases at a constant rate of 33 pounds per day (linear model), with no changes in conditions and no pauses in melting.
  • 14. Caleb and James age functions
  • a. Caleb as a function of James: C=C(J)=J+4C = C(J) = J + 4
  • b. James as a function of Caleb: J=J(C)=C4J = J(C) = C – 4
  • c. Part (a) variables: Dependent: CC; Independent: JJ
  • d. Part (b) variables: Dependent: JJ; Independent: C

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