Question

1. use the mnemonic \oscar had a hold on arthur\ as an aid in writing the definitions of sine a, cosine a, and tangent a.
2. use a calculator to evaluate: (a) $417\cos51.5^{\circ}$ (b) $32.6\tan86.3^{\circ}$
3. use the definitions of sine, cosine, and tangent and the triangles shown to find to two decimal places: (a) $\sin a$, (b) $\cos b$, and (c) $\tan c$.

(there are three right - triangle figures here, the first with legs 5 and 5.74, hypotenuse 7.6; the second with legs 7 and 6, hypotenuse 9.2; the third with legs 6 and 8, hypotenuse 10)

1. use unit multipliers to convert $4\text{ ft}^3$ to cubic inches.
2. find $b$: $\frac{a}{b}-x = \frac{p}{z}$
3. find $p$: $\frac{xz}{p}-k = \frac{m}{c}$
4. find $m$: $\frac{x}{m}-\frac{k}{c}=\frac{p}{z}$
5. solve for $a$, $b$, and $c$.

(there is a right - triangle - related figure with segments 4, 2, and $\frac{7}{2}$)
solve by factoring:

1. $-x^3 - 9x^2 = 20x$
2. $4x^2 - 81 = 0$
3. expand: $(x - 3)^3$

simplify:

1. $\frac{5x - 6x^2 + x^3}{x^2 + 2x - 3}\cdot\frac{x^2 - 9}{x^3 - 8x^2 + 15x}$
2. $16^{-3/4}$
3. $\frac{4x + \frac{1}{x}}{\frac{ay^2}{x}-4}$
4. $\sqrt{\frac{3}{7}}-4\sqrt{\frac{7}{3}}$
5. $3\sqrt{2}(5\sqrt{12}-\sqrt{2})$
6. the altitude of this isosceles triangle is 4. find $a$ and $b$.

(there is an isosceles - triangle figure with base 4)

1. find the equation of the line that passes through the point $(-2, -5)$ and has a slope of $-\frac{1}{7}$.
2. solve: $\frac{5x - 7}{2}-\frac{3x - 2}{5}=4$
3. estimate: $\frac{(47,816\times10^{5})(4923\times10^{-14})}{403,000}$

Explanation:

Step 1: Evaluate 417 cos 51.5°

Ensure calculator is in degree mode:
$417 \cos(51.5^\circ) \approx 417 \times 0.6225 = 259.58$

Step 2: Evaluate 32.6 tan 86.3°

$32.6 \tan(86.3^\circ) \approx 32.6 \times 15.89 = 518.01$

Step 3: Calculate sin A (triangle 1)

$\sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{5}{7.6} \approx 0.66$

Step 4: Calculate cos B (triangle 2)

$\cos B = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{7}{9.2} \approx 0.76$

Step 5: Calculate tan C (triangle 3)

$\tan C = \frac{\text{opposite}}{\text{adjacent}} = \frac{6}{8} = 0.75$

Step 6: Convert ft³ to in³

1 ft = 12 in, so $1\ \text{ft}^3 = 12^3\ \text{in}^3 = 1728\ \text{in}^3$
$4\ \text{ft}^3 = 4 \times 1728 = 6912\ \text{in}^3$

Step 7: Solve for b (Q10)

Rearrange $\frac{a}{b} - x = \frac{p}{z}$:
$\frac{a}{b} = \frac{p}{z} + x = \frac{p + xz}{z}$
$b = \frac{az}{p + xz}$

Step 8: Solve for p (Q11)

Rearrange $\frac{xz}{p} - k = \frac{m}{c}$:
$\frac{xz}{p} = \frac{m}{c} + k = \frac{m + kc}{c}$
$p = \frac{cxz}{m + kc}$

Step 9: Solve for m (Q12)

Rearrange $\frac{x}{m} - \frac{k}{c} = \frac{p}{z}$:
$\frac{x}{m} = \frac{p}{z} + \frac{k}{c} = \frac{pc + kz}{cz}$
$m = \frac{cxz}{pc + kz}$

Step 10: Solve for a, b, c (Q13)

Similar triangles: $\frac{a}{4+2} = \frac{4}{6} \implies a = 4$; $\frac{b}{6} = \frac{\frac{7}{2}}{4} \implies b = \frac{21}{4} = 5.25$;
$c = \sqrt{6^2 + 5.25^2} = \sqrt{36 + 27.5625} = \sqrt{63.5625} = 7.97$

Step 11: Factor -x³ -9x²=20x (Q14)

Rearrange: $-x^3 -9x^2 -20x = 0 \implies -x(x^2 +9x +20)=0$
$-x(x+4)(x+5)=0$
Solutions: $x=0, x=-4, x=-5$

Step 12: Factor 4x²-81=0 (Q15)

Difference of squares: $(2x-9)(2x+9)=0$
Solutions: $x=\frac{9}{2}, x=-\frac{9}{2}$

Step 13: Expand (x-3)³ (Q16)

Use binomial theorem:
$(x-3)^3 = x^3 - 3x^2(3) + 3x(3^2) - 3^3 = x^3 -9x^2 +27x -27$

Step 14: Simplify Q17 factor terms

Factor numerators/denominators:
$\frac{x^3-6x^2+5x}{x^2+2x-3} \cdot \frac{x^2-9}{x^3-8x^2+15x} = \frac{x(x-1)(x-5)}{(x+3)(x-1)} \cdot \frac{(x-3)(x+3)}{x(x-3)(x-5)}$
Cancel common terms: $1$

Step 15: Simplify 16^(-3/4) (Q18)

$16^{-3/4} = (2^4)^{-3/4} = 2^{-3} = \frac{1}{8}$

Step 16: Simplify Q19

Multiply numerator/denominator by x:
$\frac{4x^2 +1}{ay^2 -4x}$

Step 17: Simplify Q20 rationalize radicals

$\sqrt{\frac{3}{7}} -4\sqrt{\frac{7}{3}} = \frac{\sqrt{21}}{7} - \frac{4\sqrt{21}}{3} = \frac{3\sqrt{21}-28\sqrt{21}}{21} = -\frac{25\sqrt{21}}{21}$

Step 18: Simplify Q21 distribute

$3\sqrt{2}(5\sqrt{12}-\sqrt{2}) = 15\sqrt{24} - 3\sqrt{4} = 15 \times 2\sqrt{6} - 3 \times 2 = 30\sqrt{6} -6$

Step 19: Find A,B (Q22)

Half base = 2, altitude=4:
$\tan A = \frac{4}{2}=2 \implies A = \arctan(2) \approx 63.43^\circ$
$B = 180^\circ - 2 \times 63.43^\circ = 53.14^\circ$

Step 20: Line equation (Q23)

Point-slope form: $y - (-5) = -\frac{1}{2}(x - (-2))$
Simplify: $y +5 = -\frac{1}{2}x -1 \implies y = -\frac{1}{2}x -6$

Step 21: Solve Q24 clear denominators

Multiply by 10: $5(5x-7) -2(3x-2)=40$
$25x-35-6x+4=40$
$19x -31=40 \implies 19x=71 \implies x=\frac{71}{19} \approx 3.74$

Step 22: Estimate Q25

Round values: $47.816 \times10^5 \approx 4.8 \times10^6$, $4923 \times10^{-14} \approx 5 \times10^{-11}$, $403000 \approx 4 \times10^5$
$\frac{(4.8 \times10^6)(5 \times10^{-11})}{4 \times10^5} = \frac{24 \times10^{-5}}{4 \times10^5} = 6 \times10^{-10}$

Answer:

1. (a) $\boldsymbol{259.58}$; (b) $\boldsymbol{518.01}$
2. (a) $\boldsymbol{0.66}$; (b) $\boldsymbol{0.76}$; (c) $\boldsymbol{0.75}$
3. $\boldsymbol{6912}$ cubic inches
4. $\boldsymbol{b = \frac{az}{p + xz}}$
5. $\boldsymbol{p = \frac{cxz}{m + kc}}$
6. $\boldsymbol{m = \frac{cxz}{pc + kz}}$
7. $\boldsymbol{a=4}$, $\boldsymbol{b=5.25}$, $\boldsymbol{c=7.97}$
8. $\boldsymbol{x=0,\ x=-4,\ x=-5}$
9. $\boldsymbol{x=\frac{9}{2},\ x=-\frac{9}{2}}$
10. $\boldsymbol{x^3 -9x^2 +27x -27}$
11. $\boldsymbol{1}$
12. $\boldsymbol{\frac{1}{8}}$
13. $\boldsymbol{\frac{4x^2 +1}{ay^2 -4x}}$
14. $\boldsymbol{-\frac{25\sqrt{21}}{21}}$
15. $\boldsymbol{30\sqrt{6} -6}$
16. $\boldsymbol{A \approx 63.43^\circ}$, $\boldsymbol{B \approx 53.14^\circ}$
17. $\boldsymbol{y = -\frac{1}{2}x -6}$
18. $\boldsymbol{x=\frac{71}{19} \approx 3.74}$
19. $\boldsymbol{6 \times10^{-10}}$