Grade 11 Math Practice Test Questions

a) Simplify each term individually:
$\sqrt{125} = 5\sqrt{5}$
$\frac{16}{\sqrt{32}} = \frac{16}{4\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$
$\sqrt{20} = 2\sqrt{5}$
Combine: $(5\sqrt{5} - 2\sqrt{5}) + (3\sqrt{2} - 2\sqrt{2}) = \mathbf{3\sqrt{5} + \sqrt{2}}$.

b) $5\sqrt{12} = 10\sqrt{3}$ and $2\sqrt{27} = 6\sqrt{3}$.
$10\sqrt{3}(-6\sqrt{3} - 6\sqrt{3}) = 10\sqrt{3}(-12\sqrt{3}) = -120(3) = \mathbf{-360}$.

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Rationalize the denominator by multiplying by the conjugate $(2 + \sqrt{5})$:
\[ \frac{(1+\sqrt{5})(2+\sqrt{5})}{4-5} = \frac{2 + \sqrt{5} + 2\sqrt{5} + 5}{-1} = \frac{7+3\sqrt{5}}{-1} = -7-3\sqrt{5} \]
Simplify $4\sqrt{20} = 8\sqrt{5}$.
Final result: $-7 - 3\sqrt{5} + 8\sqrt{5} = \mathbf{-7 + 5\sqrt{5}}$.

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Distribute the first part: $x^3 - 2x^2 + 2x - 3x^2 + 6x - 6 = x^3 - 5x^2 + 8x - 6$.
Expand the second part: $(x+3)^2 = x^2 + 6x + 9$.
Subtract: $(x^3 - 5x^2 + 8x - 6) - (x^2 + 6x + 9) = \mathbf{x^3 - 6x^2 + 2x - 15}$.

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Since $b > 0$, $\sqrt{9b^2} = 3b$.
Equation: $3b = 6b\sqrt{x}$.
Divide by $6b$: $\frac{3b}{6b} = \sqrt{x} \implies \frac{1}{2} = \sqrt{x}$.
Square both sides: $\mathbf{x = 1/4}$.

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a) Common denominator is $2(x-1)(x+2)$.
Numerator: $2x(2)(x+2) - 1(2)(x-1) - 6 = 4x^2 + 8x - 2x + 2 - 6 = 4x^2 + 6x - 4 = 2(2x-1)(x+2)$.
Simplify: $\mathbf{\frac{2x-1}{x-1}}$.

b) Factor all terms and multiply by the reciprocal:
\[ \frac{2(x+2)(x-1)}{(x+3)(x+5)} \cdot \frac{(x+7)(x+3)}{2(x+1)(x+2)} = \mathbf{\frac{(x-1)(x+7)}{(x+5)(x+1)}} \]

Watch 5a | Watch 5b

Simplify first part: $\frac{9x^4y^4}{16x^4y^8} = \frac{9}{16y^4}$.
Simplify second part: $\frac{27x^3y^3}{36x^{-2}y^4} = \frac{3x^5}{4y}$.
Divide: $\frac{9}{16y^4} \cdot \frac{4y}{3x^5} = \mathbf{\frac{3}{4x^5y^3}}$.

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Rewrite as $x^2 + 2x - 2 < 0$.
Solve the equality $x^2 + 2x - 2 = 0$ using the quadratic formula to find critical points: $x = -1 \pm \sqrt{3}$.
Since the parabola opens upward ($a > 0$), the expression is negative between the roots: $\mathbf{(-1-\sqrt{3}, -1+\sqrt{3})}$.

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a) Vertex: $x = -b/2a = -2/-2 = 1$. $f(1) = 3$. Vertex is $\mathbf{(1, 3)}$.
b) Intercepts: y-intercept at $f(0) = 2$. x-intercepts at $\mathbf{1 \pm \sqrt{3}}$.
c) Symmetry: $x = 1$.
d) Domain: All real numbers. Range: $(-\infty, 3]$.

Use vertex form: $y = a(x-2)^2 + 2$.
Plug in y-intercept $(0, -2)$: $-2 = a(0-2)^2 + 2 \implies -4 = 4a \implies a = -1$.
Expand: $f(x) = -(x^2 - 4x + 4) + 2 = \mathbf{-x^2 + 4x - 2}$.
Exact x-intercepts ($y=0$): $\mathbf{x = 2 \pm \sqrt{2}}$.

Using $(1, 1)$: $1 = a^{1-b}$. For any $a > 0$, this implies $1-b = 0$, so $\mathbf{b = 1}$.
Using $(2, 3)$: $3 = a^{2-1} \implies a^1 = 3$, so $\mathbf{a = 3}$.
Equation: $\mathbf{g(x) = 3^{x-1}}$.

Multiply by LCD $(x-2)(x+1)$:
$(2x+1)(x+1) = -1(x-2)(x+1) - 1(x-2)$
$2x^2 + 3x + 1 = -(x^2 - x - 2) - x + 2$
$2x^2 + 3x + 1 = -x^2 + 4 \implies 3x^2 + 3x - 3 = 0$.
Divide by 3: $x^2 + x - 1 = 0$. Solution: $\mathbf{x = \frac{-1 \pm \sqrt{5}}{2}}$.

a) $\cos(45+30) = \cos(45)\cos(30) - \sin(45)\sin(30) = \mathbf{\frac{\sqrt{6}-\sqrt{2}}{4}}$.
b) $\sec(15) = 1/\cos(45-30) = 1/(\cos45\cos30 + \sin45\sin30) = \mathbf{\frac{4}{\sqrt{6}+\sqrt{2}}}$.

a) $\tan(-330^{\circ}) = \tan(30^{\circ}) = \mathbf{\frac{\sqrt{3}}{3}}$.
b) $\csc(480^{\circ}) = \csc(120^{\circ}) = 1/\sin(60^{\circ}) = \mathbf{\frac{2\sqrt{3}}{3}}$.

LHS: $\tan x \cdot \frac{1}{\tan x} + \frac{1}{\cos x} \cdot \sin x$
$= 1 + \frac{\sin x}{\cos x}$
$= 1 + \tan x$. RHS achieved.

a) $\theta = \arctan(0.2) \approx 11.3^{\circ}$. Also a solution in Quarter 3: $11.3 + 180 = \mathbf{191.3^{\circ}}$.
b) $\theta + 30 = 60^{\circ}$ or $300^{\circ} \implies \theta = \mathbf{30^{\circ}, 270^{\circ}}$.

a) Max Depth: $7.2 + 5.8 = \mathbf{13.0 \text{ m}}$. Occurs when $\cos = 1$: $t - 6.5 = 0 \implies \mathbf{t=6.5}$.
b) Min Depth: $-7.2 + 5.8 = \mathbf{-1.4 \text{ m}}$. Occurs when $\cos = -1$: $30(t-6.5) = 180 \implies t = 12.5 $.

Parent function $2^x$. Reflect across x-axis ($-2^x$), shift right 2, shift down 3. Asymptote at $y = -3$.

Amplitude: 3. Midline: $y=3$. Reflection across midline. Phase shift: $30^{\circ}$ right.

Multiply coefficients: $2^{1/5} \cdot (16)^{1/5} = 2^{1/5} \cdot (2^4)^{1/5} = 2^{1/5} \cdot 2^{4/5} = 2^1 = 2$.
Multiply variables: $x^{1/2} \cdot x^{1/2} = x^1 = x$.
Final result: $\mathbf{2x}$.

Express $8 ^x \; 4^x$ with base 2: $8 ^x \; 4^x = 2^{3x} \cdot 2^{2x} = 2^{5x} \implies \dfrac{1}{8 ^x \; 4^x} = 2^{-5x} $.
Equate exponents: $-5x = -7x + 0.5 \implies 2x = 0.5 \implies \mathbf{x = 0.25}$.

Standard form: $2x^2 - x + (m-1) = 0$.
Two real solutions require $D = b^2 - 4ac > 0$.
$(-1)^2 - 4(2)(m-1) > 0 \implies 1 - 8m + 8 > 0 \implies 9 > 8m \implies \mathbf{m < 9/8}$.

Solve line for $y$: $2y = 4x - 4 \implies y = 2x - 2$.
Substitute: $(x-2)^2 + (2x-2+1)^2 = 5 \implies (x-2)^2 + (2x-1)^2 = 5$.
Expand: $5x^2 - 8x + 5 = 5 \implies 5x^2 - 8x = 0$.
$x(5x-8)=0 \implies x=0, x=1.6$. Points: $\mathbf{(0, -2)}$ and $\mathbf{(1.6, 1.2)}$.

Ratio $r = 54/-18 = -3$. First term $a = -18/(-3)^2 = -2$.
$a_7 = -2(-3)^6 = \mathbf{-1458}$.
$S_{10} = \frac{-2(1 - (-3)^{10})}{1 - (-3)} = \frac{-2(-59048)}{4} = \mathbf{29524}$.

Use formula $PV = A(1 + r/n)^{-nt}$.
$PV = 20000(1 + 0.03)^{-20} = \mathbf{\$11,073.52}$.