MATH 130 · Wednesday, October 28 · Test 2 review

Test 2 is Friday. Today: every problem type on it.

  • Turn in Quiz 9 now.
  • Friday, October 30: 15 questions, 200 points, 4 pages, 50 minutes.
  • Every question is a skill from Quizzes 5 to 9.
  • For each type, I work a model; then you try one on paper before the answer appears.

Where the 200 points are

  • Logarithms (Section 4.3): 60
  • Angles and polygons: 60
  • Circles: 40
  • Interest (Section 4.2): 20
  • Degrees, minutes, seconds: 20
Test 2 review
Test 2 · what is on each page

The test, question by question

QSkillPts
1Compound interest10
2Continuous compounding10
3Exponential → logarithmic form15
4Logarithmic → exponential form15
5Solve ln x = c15
6Solve bx = c (change of base)15
7Parallel lines: equation, x, angle20
8Similar triangles: angle and side20
QSkillPts
9Trapezoid area (a slanted side is given)10
10Heron's formula10
11Decimal degrees → DMS10
12DMS → decimal degrees10
13Inscribed angle and arc length20
14Sector area10
15Linear speed of a wheel10
Highlighted: new or changed this week, worth 70 points together.
Test 2 review
Test 2 · formulas

Printed on the test, or from memory?

Printed on Test 2

  • A = P(1 + r/n)nt
  • A = Pert
  • logb a = ln a ÷ ln b change of base
  • A = √(s(s − a)(s − b)(s − c)), s = (a + b + c)/2

From memory

  • by = x ⟺ logb x = y; ln x = c ⟺ x = ec
  • trapezoid: A = ½(b₁ + b₂)h
  • DMS: D + M/60 + S/3600
  • inscribed angle = ½ arc
  • radians = degrees × π/180; s = rθ; A = ½r²θ
  • v = rω, with ω = rpm × 2π
Test 2 review
Section 4.2 · Question 1 · Compound interest (Quiz 5)

A = P(1 + r/n)nt

r as a decimal · the test gives n: monthly (n = 12), quarterly (n = 4), semiannually (n = 2) · round to the nearest cent at the end

Model

$2,500 at 3.5% compounded monthly (n = 12) for 6 years

A = 2500(1 + 0.035/12)12·6 = 2500(1 + 0.035/12)72

A ≈ $3,083.25

You try

$7,000 at 5.2% compounded quarterly (n = 4) for 9 years

A = 7000(1 + 0.052/4)36

A ≈ $11,143.92

Test 2 review
Section 4.2 · Question 2 · Continuous compounding (Quiz 5)

A = Pert

"compounded continuously" means e, with no n · multiply r × t first · round to the nearest cent

Model

$1,500 at 6.2% compounded continuously for 7 years

A = 1500e0.062·7 = 1500e0.434

A ≈ $2,315.13

You try

$9,000 at 3.8% compounded continuously for 11 years

A = 9000e0.418

A ≈ $13,670.29

Test 2 review
Section 4.3 · Questions 3 and 4 · Exponential and logarithmic form (Quiz 6)

The log is the exponent, even a fraction

by = x  ⟺  logb x = y
253/2= 125 log25125 = 3/2
Model

(a) Write 43/2 = 8 in log form.  (b) Write log9 27 = 1.5 in exponential form.

(a) log4 8 = 3/2

(b) 91.5 = 27

You try

(a) Write 253/2 = 125 in log form.  (b) Write log4 32 = 2.5 in exponential form.

(a) log25 125 = 3/2

(b) 42.5 = 32

Test 2 review
Section 4.3 · Questions 5 and 6 · Solving with logarithms (Quiz 6)

ln x = c and bx = c

ln x = c → x = ec  ·  bx = c → x = logb c = ln c ÷ ln b change of base is printed  ·  four decimal places

Model

Solve. (a) ln x = 2.5  (b) 3x = 50

(a) x = e2.5 ≈ 12.1825

(b) x = log3 50 = ln 50 ÷ ln 3 ≈ 3.5609

You try

Solve. (a) ln x = 4.2  (b) 6x = 150

(a) x = e4.2 ≈ 66.6863

(b) x = ln 150 ÷ ln 6 ≈ 2.7965

Test 2 review
Geometry · Question 7 · Parallel lines and a transversal (Quiz 7)

Equal or supplementary?

Equal: alternate interior (1 and 2), alternate exterior (3 and 4), corresponding.  Add to 180°: same-side interior. Then substitute x back to get the angle.

1234 mn
Model

Alternate interior angles: (2x + 27)° and (4x − 13)°

2x + 27 = 4x − 13 → x = 20

angle = 2(20) + 27 = 67°

You try

Same-side interior angles: (2x + 20)° and (4x + 10)°

(2x + 20) + (4x + 10) = 180 → x = 25

angles 70° and 110°

Test 2 review
Geometry · Question 8 · Similar triangles (Quiz 7)

Match letters in order: A↔D, B↔E, C↔F

Matching angles are equal. Matching sides share one scale factor: keep one triangle on top, AB/BC = DE/EF.

ABC DEF 610 38°75° 9?
Model

∠A = 38°, ∠B = 75°; AB = 6, BC = 10, DE = 9. Find ∠F and EF.

∠F = ∠C = 180 − 38 − 75 = 67°

6/10 = 9/EF → 6·EF = 90 → EF = 15

You try

∠A = 49°, ∠C = 76°; AB = 14, BC = 21, DE = 12. Find ∠E and EF.

∠E = ∠B = 180 − 49 − 76 = 55°

14/21 = 12/EF → EF = 18

Test 2 review
Geometry · Question 9 · Trapezoid area (Quiz 7)

Trapezoid: use the height, not the slanted side

A = ½(b₁ + b₂)h, from memory. h is the perpendicular height (the dashed segment). The test also gives a slanted side: it is not used.

10 m18 m h = 7 m8 m
Model

Parallel sides 10 m and 18 m, slanted side 8 m, height 7 m

A = ½(10 + 18)(7) = 98 m²

The 8 m slanted side is not used.

You try

Parallel sides 6.5 cm and 11.5 cm, slanted side 5 cm, height 4 cm

A = ½(6.5 + 11.5)(4) = 36 cm²

Test 2 review
Geometry · Question 10 · Heron's formula (Quiz 7)

Three sides: semiperimeter first

printed: s = (a + b + c)/2, then A = √(s(s − a)(s − b)(s − c)) · half the perimeter, not all of it

Model

A triangle has sides 5, 12 and 13. Find its area.

s = (5 + 12 + 13)/2 = 15

A = √(15 · 10 · 3 · 2) = √900 = 30

You try

A triangle has sides 7, 15 and 20. Find its area.

s = 21

A = √(21 · 14 · 6 · 1) = √1764 = 42

Test 2 review
Geometry · Questions 11 and 12 · Degrees, minutes, seconds (Quiz 8)

DMS both ways

decimal → DMS: × 60, then × 60 again (nearest second) · DMS → decimal: D + M/60 + S/3600 (four places) · show the conversion

Model

(a) 54.2875° to DMS  (b) 31°12′45″ to decimal degrees

(a) 0.2875 × 60 = 17.25′; 0.25 × 60 = 15″ → 54°17′15″

(b) 31 + 12/60 + 45/3600 = 31.2125°

You try

(a) 98.4561° to DMS  (b) 17°53′24″ to decimal degrees

(a) 98°27′22″

(b) 17.8900°

Test 2 review
Geometry · Question 13 · Inscribed angle and arc length (Quiz 9)

Half the arc; radians before s = rθ

central angle AOC = arc AC · inscribed angle ABC = ½ arc · radians = degrees × π/180 · s = rθ with θ in radians

BACO 120°r = 8
Model

Radius 8, central angle AOC = 120°. Find ∠ABC and arc AC.

∠ABC = 120 ÷ 2 = 60°

120 × π/180 = 2π/3; s = 8(2π/3) = 16π/3 ≈ 16.76

You try

Radius 10, central angle AOC = 84°. Find ∠ABC and arc AC.

∠ABC = 42°

84° = 7π/15; s = 14π/3 ≈ 14.66

Test 2 review
Geometry · Questions 14 and 15 · Sector area and linear speed (Quiz 9)

A = ½r²θ, and v = rω

θ in radians · a diameter? halve it for r · ω = rpm × 2π rad per minute · v = rω in ft/min

r = 10θ = 2π/5
Model

Sector r = 10, θ = 2π/5. A wheel 5 ft in diameter at 160 rpm.

A = ½(100)(2π/5) = 20π ≈ 62.83

r = 2.5; ω = 320π; v = 800π ≈ 2513.27 ft/min

You try

Sector r = 6, θ = 5π/4. A wheel 3 ft in diameter at 210 rpm.

A = 22.5π ≈ 70.69

r = 1.5; v = 630π ≈ 1979.20 ft/min

Test 2 review
Test 2 · before you hand it in

Where points are usually lost

Q1–2 Rate as a percent (5.2) instead of a decimal (0.052).
Q7 Solving for x but never finding the angle.
Q3–4 Swapping the exponent and the result in log form.
Q8 Matching sides by the picture instead of the letter order.
Q6 Change of base as ln(c ÷ b) instead of ln c ÷ ln b.
Q9 Using the slanted side as the height.
Q5–6 Rounding in the middle; round to four places only at the end.
Q10 Using the full perimeter instead of s.
Q12 Dividing seconds by 60 instead of 3600.
Q13 Doubling the arc instead of halving it, or degrees in s = rθ.
Q11–12 An answer with no conversion shown.
Q15 Using the diameter as the radius, or leaving ω in rpm.
Test 2 review
Test 2 · Friday, October 30

Before Friday

Tonight and Thursday

  • Cover the "from memory" formulas and write them until you can without looking.
  • Redo every "you try" from today without notes.
  • Work the Test 2 practice PDF on the course page, then check it.
  • Take the Test 2 readiness check on the review page until you reach 80%.

Friday, in class

  • 15 questions, 200 points, 50 minutes, on paper.
  • Printed: both interest formulas, change of base and Heron's formula.
  • Bring a scientific calculator.
  • Show your work inside each question's box.
Test 2 review
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