Hephaestus

\(S_1*S_2 = 80\)

\([S_1+1]*[S_2-1] = 90\)

\(S_1 = {80\over S_2}\)

\([{80\over S_2}+1]*[S_2-1]=90\)

\([{80+S_2\over S_2}]*[{S_2-1\over 1}]=90\)

\({{80S_2-80+(S_2)^2-S_2}\over S_2}=90\)

\({{80S_2-80+(S_2)^2-S_2}} =90S_2\)

\({{80S_2-80+(S_2)^2-S_2-90S_2}} =0\)

\({-80+(S_2)^2-11S_2} =0\)

\((S_2)^2 -11S_2-80 =0\)

\(S_2=16\)

\(S_1 = 5\)

\(p = 42\)

_{graph 1:}

\(V= {[(15)5]22}+[(5^2)22]\)

\(V = 2200\)

_{graph 2:}

\((8*14*5)+({8\over2}^2)\pi \)

\(V = 1263.7\)

\(ratio_1= 11:24\)

\(class={school\over 20}\)

\(school = {class[20]}\)

\(ratio_1 = 220:480\)

_Total:

\(220\)

\(5,688,286,422\)

\(T_b = T_s[60]-T_g[21]\)

\(T_b = 39\)

\(J_b= T_b[39]-S_b[20]\)

\(J_b = 19\)

\(J_g = T_j[34]-J_b[19]\)

\(J_g= 15\)

_Answer:

\(15\)

\({-{1\over 2}}= {{1\over 2}\over{1-{1\over 2}}}\)

₌

\(-{3\over 2}\)

\(A_Q ={1\over5}AB\)

\(Arc[AP]= {1\over 5}Arc[AB]\)

\(PQ = { \sqrt{(P_O)​​^2-(AB)​​^2}}\)

\(PQ = 4\)

\(PB = {\sqrt{​​(PQ)^2+(QB)^2} }\)

\(PB =\sqrt{80}\)

_R:

\(PB = {4\sqrt{5}}\)

_If \(FG\) ₌ \(FR\), _Arc \(FG\) _{= Arc} \(FR\).    

\(FQ = 28°\) 

_Total:

_{\(360-[2(130°)+28°]= \angle RPG\)}

_{\(\angle RPG = 72\)}

\({(x+y)\over 2} =5\)

\( \sqrt{xy} =3\)

\(xy = 9\)

\(x = {9\over y}\)

\({({{9\over y}+y)}\over 2} = 5\)

\({({{9\over y}+y)}} =10\)

\({({{9\over y}+{y^2\over y})}} =10\)

\({9+y^2} = 10y\)

\((y-9)(y-1) = 0\)

\(y = 1\)

\(x = 9\)

\(y = 9\)

\(x = 1\)

\(h_e = T(3)C_e(2)\)

_{Singular hen:}

\(h= {E_1\over 1.5(12H)}\)

_{Total between 4 hens:}

\(Total_i = {24\over C_h(4)} \)

_{Individual threshold:}

\(6\)

_{Time for individual completion:}

\(6=3_d({I_t\over C_e[2]})\)

_Answer:

\(9\space days\)

_Time(Hours):    \(  1        \space  \space  \space   \space    \space    \space   2    \space   \space       \space    \space    \space  3    \space      \space    \space    \space    \space    4    \space     \space  \space    \space    \space     5    \space    \space    \space   \space    \space     6    \space   \space   \space    \space    \space     7    \space   \space    \space    \space    \space     8    \space    \space    \space    \space    \space     9      \space  \space     \space    \space    \space   10    \space     \space    \space    \space    11 \)

_{Sandra(km):       \( 90\space \space \space  180\space\space\space    270\space\space\space    360\space\space\space    450\space\space\space   540\space\space   630 \space\space\space   720  \space\space\space  810\space\space\space    900 \space\space\space    990\)}

_{Paul(km):                           \(\space \space \space110   \space \space   220   \space \space 330 \space \space \space   440   \space \space \space  550 \space \space \space    660   \space \space \space   770   \space \space \space  880  \space \space \space    990\)}

\(x(90)= [x-2](110)\)

_{Answer: \(11\space hours\)}

\(6n = 36\)

_{positive divisors:}

\(1, 2, 3, 4, 6, 9, 12, 18,36\)

_therefore:

\(n = 6\)

\(6x +2y= 40\)

\(-7x -2y = -44\)

_add:

\(6x-7x= 40-44\)

\(-x = -4\)

_result:

\(x = 4\)

\(y = 8\)

\(x = {v*t}\)

V_{i = initial velocity}

T_{f = second time}

Ti _{= first time}

\(10= {Vi* Ti}\)

\(10 = {(Vi-{3km\over h}(tf))}\)

\(Vi*Ti= {(Vi- { 3km\over h}})*tf\)

\(Tf = {7h-Ti}\)

\(Vi*Ti = (Vi-{3km\over h})*(7h-Ti)\)

\(Vi*Ti = ({Vi*7h-ViTi-{3km\over h}}+{3km \over h}*Ti)\)

\(Vi*Ti= {7h*Vi - 21km + {3km\over h}}*Ti\)

\(20km = {7h*Vi - 21km +{ 3km\over h}}*Ti\)

\(41km ={ {70km*h+ { 3km\over h}}*Ti^2\over Ti}\)

\(41km ={ {7h*10km \over Ti}+{3km*Ti\over h}\over Ti}\)

\(0={ {3km*Ti^2\over h} - 41km*Ti+70 km*h }\)

\(0 = {3x^2 -41 x + 70 }\)

_{This equation gives us two answers:}

_{2 = the number of hours in the initial trip}

_{How fast did she walk going out to the country?}

\(5km\over h\)

_Cases:

_{No repetition of 2:}

\(cases = {5!}\)

_{2 is present two times:}

\(cases = {{5!} \over 2!}\)

_{2 is present three time:}

\(cases = {{5!} \over 3!}\)

_{2 is present four time:}

\(cases = {{5!} \over 4!}\)

_{2 is present five time:}

\(cases = {{5!} \over 5!}\)

_{all cases:}

\(total = {5!+{{5!} \over 2!}+{{5!} \over 3!}+{{5!} \over 4!}+{{5!} \over 5!}}\)

\(total = {206}\)

Ratio: \(Ratio= 3:7\)

a)  _{White, because there are more white candies that the other colours:}

_{2 red, 2 green , 3 white.}

b) \(probability= {{3} \over 7}\)

    \(Ratio= 3:7\)

    \(percentage = 42.857\)

c) \(probability= {{4} \over 7}\)

    \(percentage = {57.1428}\)

Volume of a pyramid:

\(V = {area{*height} \over 3}\frac{}{}\)

Area of base = _{16 inches}

Height =

\(h =\sqrt{4^2- ({(\sqrt{16+16})​​ \over 2})^2}{}_{}\)

\(h= {\sqrt{8}}\)

Final formula:

\(V= {1\over 3}* 16 * \sqrt{8}\)

\(V = 15.0849446653\)

\(V = 15.08\)

Volume = 15.08