Factor: \(x^2 - 7x + 12 = (x-3)(x-4) = 0\)
We need two numbers that multiply to +12 and add to −7.
→ Those are −3 and −4... wait! (−3)×(−4) = +12 ✓ but (−3)+(−4) = −7 ✓
But the factors are \((x-3)(x-4)\), giving \(x = 3\) or \(x = 4\).
Trick: signs flip when you set each factor = 0!

\(a=2, b=-4, c=3\)
Discriminant: \(\Delta = b^2 - 4ac = (-4)^2 - 4(2)(3) = 16 - 24 = -8\)
Since \(\Delta < 0\), there are no real solutions. ✓
Remember: negative discriminant → no real roots (complex only)

\(x^2 + 6x + 5\)
Half of 6 = 3, so try \((x+3)^2 = x^2 + 6x + 9\)
But we only have +5, so: \(x^2+6x+5 = (x+3)^2 - 9 + 5 = (x+3)^2 - 4\)
Therefore \(b = -4\) ✓
Key: add (b/2)², then subtract it again!

\(g(x) = (x+4)^2 - 1\) is in vertex form \((x-h)^2 + k\)
Here \(h = -4\) (because \(x+4 = x-(-4)\)) and \(k = -1\)
Vertex = \((h, k) = (-4, -1)\) ✓
Inside parentheses: OPPOSITE sign for h!

Two restrictions:
1. Square root: \(x - 3 \geq 0 \Rightarrow x \geq 3\)
2. Denominator ≠ 0: \(\sqrt{x-3} \neq 0 \Rightarrow x \neq 3\)
Combining: \(x > 3\) ✓
Square root needs ≥0, but denominator needs ≠0. Combine!

Work inside out:
Step 1: \(g(3) = 3^2 = 9\)
Step 2: \(f(g(3)) = f(9) = 2(9) + 1 = 19\) ✓
f(g(x)): do g FIRST, then f. Inside → Outside!

\(\sin(45°) = \dfrac{\text{opposite}}{\text{hypotenuse}}\)
\(\dfrac{\sqrt{2}}{2} = \dfrac{\text{opposite}}{8\sqrt{2}}\)
Opposite \(= 8\sqrt{2} \times \dfrac{\sqrt{2}}{2} = \dfrac{8 \times 2}{2} = 8\) ✓

The perpendicular from center to chord bisects it → right triangle:
\(r^2 = d^2 + (\text{half-chord})^2\)
\(10^2 = 8^2 + h^2 \Rightarrow 100 = 64 + h^2 \Rightarrow h^2 = 36 \Rightarrow h = 6\)
Full chord = \(2 \times 6 = 12\) cm ✓
Always double the half-chord for the full length!

Cosine Rule: \(c^2 = a^2 + b^2 - 2ab\cos C\)
\(= 7^2 + 5^2 - 2(7)(5)\cos 60°\)
\(= 49 + 25 - 70 \times 0.5\)
\(= 74 - 35 = 39\) ✓

Area \(= \dfrac{1}{2}ab\sin C = \dfrac{1}{2} \times 6 \times 8 \times \sin 30°\)
\(= \dfrac{1}{2} \times 48 \times 0.5 = 12\) ✓
sin 30° = 0.5 — memorize this!

From term 3 to term 7: 4 steps of \(d\)
\(u_7 - u_3 = 4d \Rightarrow 27 - 11 = 4d \Rightarrow 16 = 4d \Rightarrow d = 4\) ✓

Terms: 2, 6, 18, 54
\(S_4 = \dfrac{2(3^4 - 1)}{3-1} = \dfrac{2(81-1)}{2} = \dfrac{2 \times 80}{2} = 80\) ✓
Or just: \(2 + 6 + 18 + 54 = 80\)

\(u_1 = 3, d = 4\)
\(u_{10} = 3 + 9(4) = 3 + 36 = 39\)
\(S_{10} = \dfrac{10}{2}(3 + 39) = 5 \times 42 = 210\) ✓

Ordered: 4, 7, 7, 8, 12, 15 (6 values → two middle values)
Middle values: 7 and 8
Median \(= \dfrac{7+8}{2} = 7.5\) ✓
Even count: always average the two middle values!

Ways to get exactly 2 heads: HHT, HTH, THH → 3 arrangements
Each has probability \(\left(\dfrac{1}{2}\right)^3 = \dfrac{1}{8}\)
Total: \(3 \times \dfrac{1}{8} = \dfrac{3}{8}\) ✓
Or use binomial: \(\binom{3}{2}\left(\dfrac{1}{2}\right)^2\left(\dfrac{1}{2}\right)^1 = 3 \times \dfrac{1}{8} = \dfrac{3}{8}\)

Set A has ALL SAME values → zero spread → \(\sigma = 0\)
Set B varies widely → \(\sigma > 0\)
So Set A has SMALLER (in fact, zero) standard deviation ✓
SD = 0 means no spread. All values identical!

\(\log_2 32 - \log_2 4 = \log_2\!\left(\dfrac{32}{4}\right) = \log_2 8 = \log_2 2^3 = 3\) ✓
log(a) − log(b) = log(a/b). Division law!

\(3^{2x-1} = 27 = 3^3\)
Same base: \(2x - 1 = 3\)
\(2x = 4 \Rightarrow x = 2\) ✓
Same base → equate the exponents directly!

\(P = 500 \times 2^{15/5} = 500 \times 2^3 = 500 \times 8 = 4000\) ✓
In 15 years: 3 doublings (5→10→15 years)
\(500 \to 1000 \to 2000 \to 4000\) ✓

Method 1 (Change of base): \(\log_4 64 = \dfrac{\log 64}{\log 4} = \dfrac{\log 4^3}{\log 4} = \dfrac{3\log 4}{\log 4} = 3\)
Method 2 (Direct): \(4^? = 64 \Rightarrow 4^3 = 64\) ✓
Answer: 3 ✓