heureka

how do you find these out?

i hope seeing how it works will help me apply it in similar questions in the future..

5.

Let AP = x

$\begin{array}{|rcll|} \hline \underbrace{112}_{\text{area of the shaded figue PQRS}} &=& \underbrace{16 \cdot 12}_{\text{area rectangle ABCD}} - \underbrace{2\cdot \frac{(12-x)\cdot x}{2} - 2\cdot \frac{(16-x)\cdot x}{2}}_{\text{area triangles SAP, PBQ, QCR, RDS}} \\\\ 112 &=& 192 - (12-x)\cdot x - (16-x)\cdot x \\ 112 &=& 192 - (12x-x^2) - (16x-x^2)\\ 112 &=& 192 - 12x+x^2 - 16x+x^2\\ 112 &=& 192 - 12x+x^2 - 16x+x^2\\ 112 &=& 192 - 28x + 2x^2\\ 0 &=& 192 -112 - 28x + 2x^2\\ 0 &=& 80 - 28x + 2x^2\\ 2x^2 - 28x + 80 &=& 0 \quad &| \quad :2 \\ x^2 - 14x + 40 &=& 0 \\ x &=& \frac{ 14\pm\sqrt{14^2-4\cdot 40} } {2} \\ x &=& \frac{ 14\pm\sqrt{36} } {2} \\ x &=& \frac{ 14\pm 6} {2} \\ x = \frac{ 14 + 6} {2} & \text{or} & x = \frac{ 14 - 6} {2} \\ x = \frac{ 20} {2} & \text{or} & x = \frac{ 8 } {2} \\ \mathbf{x = 10} & \text{or} & \mathbf{x = 4}\\ \hline \end{array}$

The length of AP is 10 cm or 4 cm

viewing angle 41.8 x 80' + 5'5"= ?

$\begin{array}{|rcll|} \hline && 41.8 \times 80' + 5'5" \\ &=& 41.8 \times 80' + 5' + 5" \\ &= & 41.8 \times 80' + 5'+ 5"\cdot \frac{1'}{60"} \\ &= & 41.8 \times 80' + 5'+ \left(\frac{5}{60} \right)' \\ &= & 3344' + 5'+ \left(\frac{5}{60} \right)' \\ &= & 3349' + \left(\frac{5}{60} \right)' \\ &= & 3349'\cdot \frac{60}{60} + \left(\frac{5}{60} \right)' \\ &= & \frac{3349' \cdot 60+5'}{60} \\ &= & \frac{200945'}{60} \cdot \frac{ 1^{\circ} }{60'}\\ &= & \left(\frac{200945}{60\cdot 60} \right)^{\circ}\\ &=& 55.8180\bar5 ^{\circ} \\ \hline \end{array}$

find all the fourth roots of -8 in Cartesian form.

simplify as much as possilbe

$\begin{array}{|rcll|} \hline \mathbf{z^4} & \mathbf{=} & \mathbf{-8 \qquad z_1,\ z_2,\ z_3,\ z_4 =\ ?}\\\\ z^4 &=& (-2)\cdot 4 \quad &| \quad \pm\sqrt{} \\ z^2 &=& \pm 2\cdot \sqrt{ -2 } \\ z^2 &=& \pm 2\cdot \sqrt{ (-1) \cdot 2 } \\ z^2 &=& \pm 2\cdot \sqrt{2} \cdot \sqrt{ -1 } \quad &| \quad \sqrt{ -1 } = i\\\\ \mathbf{z^2} & \mathbf{=} & \mathbf{\pm 2\cdot \sqrt{2} \cdot i }\\ \hline \end{array}$

Solutions $z_1,\ z_2,\ z_3,\ z_4$:

$\begin{array}{|lrcll|} \hline (1) & z_1^2 &=& +2\cdot \sqrt{2} \cdot i \\ (2) & z_2^2 = (-z_1)^2 &=& +2\cdot \sqrt{2} \cdot i \qquad \text{or} \qquad z_2 = -z_1\\ (3) & z_3^2 &=& -2\cdot \sqrt{2} \cdot i \\ (4) & z_4^2 = (-z_3)^2 &=& -2\cdot \sqrt{2} \cdot i \qquad \text{or} \qquad z_4 = -z_3\\ \hline \end{array}$

$\begin{array}{|lrcll|} \hline \mathbf{z_1} \\\\ (1) & z_1^2 &=& +2\cdot \sqrt{2} \cdot i \quad &| \quad z_1 =a+b\cdot i \\ & (a+b\cdot i)^2 &=& +2\cdot \sqrt{2} \cdot i \\ & a^2 + 2abi + b^2i^2 &=& +2\cdot \sqrt{2} \cdot i \quad &| \quad i^2 = -1\\ & a^2 + 2abi - b^2 &=& +2\cdot \sqrt{2} \cdot i \\ & \underbrace{a^2 - b^2}_{Re=0} + \underbrace{2abi}_{Im=+2\cdot \sqrt{2} \cdot i} &=& +2\cdot \sqrt{2} \cdot i \\\\ & a^2-b^2 &=& 0 \\ & \mathbf{a^2} & \mathbf{=} & \mathbf{b^2}\\\\ & 2abi &=& +2\sqrt{2}\cdot i \\ & ab &=& \sqrt{2} \\ & \mathbf{b} &\mathbf{=}& \mathbf{ \frac{ \sqrt{2} } {a} } \\\\ & a^2 &=& b^2 \\ & a^2 &=& (\frac{ \sqrt{2} } {a} )^2 \\ & a^2 &=& \frac{ 2 } {a^2} \\ & a^4 &=& 2 \\ & \mathbf{a} &\mathbf{=}& \mathbf{\sqrt[4]{2}} \\\\ & b &=& \frac{ \sqrt{2} } {a} \\ & b &=& \frac{ \sqrt{2} } {\sqrt[4]{2}} \\ & b &=& \frac{ 2^{\frac12} } { 2^{\frac14} } \\ & b &=& 2^{\frac12-\frac14} \\ & b &=& 2^{\frac14} \\ & \mathbf{b} &\mathbf{=}& \mathbf{\sqrt[4]{2}} \\\\ & z_1 &=& a+bi \\ & z_1 &=& \sqrt[4]{2} + \sqrt[4]{2} \cdot i \\ & \mathbf{z_1} &\mathbf{=}& \mathbf{\sqrt[4]{2}\cdot (1 + i)} \\ \hline \end{array}$

$\begin{array}{|lrcll|} \hline \mathbf{z_2} \\\\ (2) & z_2 &=& -z_1 \\ & \mathbf{z_2} &\mathbf{=}& \mathbf{-\sqrt[4]{2}\cdot (1 + i)} \\ \hline \end{array}$

$\begin{array}{|lrcll|} \hline \mathbf{z_3} \\\\ (3) & z_3^2 &=& -2\cdot \sqrt{2} \cdot i \quad &| \quad z_3 =a+b\cdot i \\ & (a+b\cdot i)^2 &=& -2\cdot \sqrt{2} \cdot i \\ & a^2 + 2abi + b^2i^2 &=& -2\cdot \sqrt{2} \cdot i \quad &| \quad i^2 = -1\\ & a^2 + 2abi - b^2 &=& -2\cdot \sqrt{2} \cdot i \\ & \underbrace{a^2 - b^2}_{Re=0} + \underbrace{2abi}_{Im=-2\cdot \sqrt{2} \cdot i} &=& -2\cdot \sqrt{2} \cdot i \\\\ & a^2-b^2 &=& 0 \\ & \mathbf{a^2} & \mathbf{=} & \mathbf{b^2}\\\\ & 2abi &=& -2\sqrt{2}\cdot i \\ & ab &=& -\sqrt{2} \\ & \mathbf{b} &\mathbf{=}& \mathbf{ \frac{ -\sqrt{2} } {a} } \\\\ & a^2 &=& b^2 \\ & a^2 &=& (\frac{ -\sqrt{2} } {a} )^2 \\ & a^2 &=& \frac{ 2 } {a^2} \\ & a^4 &=& 2 \\ & \mathbf{a} &\mathbf{=}& \mathbf{\sqrt[4]{2}} \\\\ & b &=& \frac{ -\sqrt{2} } {a} \\ & b &=& \frac{ -\sqrt{2} } {\sqrt[4]{2}} \\ & b &=& -\frac{ 2^{\frac12} } { 2^{\frac14} } \\ & b &=& -2^{\frac12-\frac14} \\ & b &=& -2^{\frac14} \\ & \mathbf{b} &\mathbf{=}& \mathbf{-\sqrt[4]{2}} \\\\ & z_3 &=& a+bi \\ & z_3 &=& \sqrt[4]{2} - \sqrt[4]{2} \cdot i \\ & \mathbf{z_3} &\mathbf{=}& \mathbf{\sqrt[4]{2}\cdot (1 - i)} \\ \hline \end{array}$

$\begin{array}{|lrcll|} \hline \mathbf{z_4} \\\\ (4) & z_4 &=& -z_3 \\ & \mathbf{z_4} &\mathbf{=}& \mathbf{-\sqrt[4]{2}\cdot (1 - i)} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \mathbf{z_1} &\mathbf{=}& \mathbf{\sqrt[4]{2}\cdot (1 + i)} \\ \mathbf{z_2} &\mathbf{=}& \mathbf{-\sqrt[4]{2}\cdot (1 + i)} \\ \mathbf{z_3} &\mathbf{=}& \mathbf{\sqrt[4]{2}\cdot (1 - i)} \\ \mathbf{z_4} &\mathbf{=}& \mathbf{-\sqrt[4]{2}\cdot (1 - i)} \\ \hline \end{array}$

The shape of your favorite kite is triangular.
The coordinates of the three vertexes of the kite are located at (4,1), (0,-8), and (-4,1).
Find the area of the region.

$\begin{array}{|rcll|} \hline \text{Area } &=& \frac12\cdot b\cdot h \\ &=& \frac12\cdot [~4-(-4)~]\cdot [~1-(-8)~] \\ &=& \frac12\cdot (4+4)\cdot (1+8) \\ &=& \frac12\cdot 8\cdot 9 \\ &=& 4\cdot 9 \\ &=& 36 \\ \hline \end{array}$

1. Domain:

$\begin{array}{|rcll|} \hline 5^{x+1}-20 &> & x 0 \quad &| \quad +20 \\ 5^{x+1} &>& 20 \quad &| \quad 5^{x+1}=5^x\cdot 5^1 \\ 5^x\cdot 5^1 &>& 20 \\ 5^x\cdot 5 &>& 4\cdot 5 \quad &| \quad : 5\\ 5^x &=& 4 \quad &| \quad \ln()\\ \ln(5^x) &>& \ln(4) \\ x\cdot \ln(5) &>& \ln(4) \quad &| \quad : \ln(5)\\ x &>& \frac{ \ln(4) } { \ln(5) } \\ x &>& \frac{ 1.38629436112 } { 1.60943791243 } \\ \mathbf{ x } & \mathbf{>} & \mathbf{ 0.86135311615 } \\ \hline \end{array}$

2. Solution:

$\begin{array}{|rcll|} \hline \log_5{5^{x+1}-20} &=& x \quad &| \quad 5^{()} \\ 5^{x+1}-20 &=& 5^x \quad &| \quad 5^{x+1}=5^x\cdot 5^1 \\ 5^x\cdot 5^1 -20 &=& 5^x \quad &| \quad +20 \\ 5^x\cdot 5 &=& 5^x +20 \quad &| \quad -5^x \\ 5^x\cdot 5 -5^x &=& 20 \\ 5^x\cdot (5-1) &=& 20 \\ 5^x\cdot 4 &=& 20 \quad &| \quad : 4 \\ 5^x &=& \frac{20}{4} \\ 5^x &=& 5 \\ 5^x &=& 5^1 \\ \mathbf{ x } & \mathbf{=} & \mathbf{1} \\ \hline \end{array}$

graph:

An airplane is flying 100 km north and 185 km west of an airport. it is flying at a height of 7 km.

what is the angle of elevation of the airplane,

from the point of view of the airport?

$\begin{array}{|rcll|} \hline \tan{\text{(angle of elevation) } } &=& \frac{7\ \text{km} } { \sqrt{ (100\ \text{km})^2 + (185\ \text{km})^2 } } \\ &=& \frac{7 } { 210.297408448 } \\ &=& 0.03328619241\\ \text{ angle of elevation } &=& \arctan(0.03328619241) \\ &=& 1.90645445031\dots ^{\circ} \\ \hline \end{array}$

The angle of elevation of the airplane from the point of view of the airport is $\approx \mathbf{1.9^{\circ}}$

2 women share 5 dozen eggs in the ratio 1:2 .

how many eggs does each get.

$\text{eggs} = 5 * 12 = 60\\ \text{Let } w_1 \text{ first women} \\ \text{Let } w_2 \text{ second women}$

$\begin{array}{|lrcll|} \hline (1) & \frac{w_1}{w_2} &=& \frac12 \quad &| \quad \cdot 2\\ & 2\cdot \frac{w_1}{w_2} &=& 1 \quad &| \quad \cdot w_2\\ & 2\cdot w_1 &=& w_2\\ & \mathbf{ w_2 } & \mathbf{=} & \mathbf{2\cdot w_1} \\\\ (2) & w_1+w_2 &=& 60 \text{ eggs} \quad &| \quad w_2 = 2\cdot w_1 \\ & w_1 +2\cdot w_1 &=& 60 \\ & 3\cdot w_1 &=& 60 \quad &| \quad : 3\\ & w_1 &=& \frac{60}{3} \\ & \mathbf{ w_1 } & \mathbf{=} & \mathbf{20 \text{ eggs} } \\\\ & w_2 & = & 60-w_1 \\ & w_2 & = & 60-20 \\ & \mathbf{ w_2 } & \mathbf{=} & \mathbf{40 \text{ eggs} } \\ \hline \end{array}$

First women get 20 eggs.

Second women get 40 eggs.

check:

20 + 40 = 60

$\frac{20}{40} = \frac12$

Navya has 10½ feet of craft wire that she uses to make earrings.

If each pair of earring requires ¾ foot of wire,

how many pairs of earrings is Navya able to make?

$\begin{array}{|rcll|} \hline && \frac{ 10+\frac12 } { \frac34 } \\\\ &=& \frac{ \frac{20}{2} +\frac12 } { \frac34 } \\\\ &=& \frac{ \frac{21}{2}} { \frac34 } \\\\ &=& \frac{21}{2} \times \frac43 \\\\ &=& \frac{21\cdot 4}{2\cdot 3} \\\\ &=& \frac{21\cdot 4}{3\cdot 2} \\\\ &=& 7\cdot 2\\ &=& 14\\ \hline \end{array}$

Navya is  able to make 14 pairs of earrings.

listing:

college class contains 62 students,

of these 32 are freshman,

32 are economics major,

and 12 are neither.

what is the probablity the student is both a freshman and economics major

$\begin{array}{|lcll|} \hline \text{Set students } ~ s &=& 62 \\ \text{Set freshman } ~ f &=& 32 \\ \text{Set economics major} ~ e_m &=& 32 \\ \text{Set neither} ~ n &=& 12 \\\\ \text{both a freshman and economics major } &=& f+e_m+n -s \\ &=& 32+32+12-62 \\ &=& 14 \\\\ \text{the probablity is the student is both a freshman and economics major } &=& \frac{f+e_m+n -s}{s} \\ &=& \frac{14}{62} \\ &=& \frac{7}{31} \quad ( 22.58\ \%)\\ \hline \end{array}$