Alebra

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3. "Diophantus passed 1/6 of his life in childhood, 1/12 in youth and 1/7 more as a bachelor. Five years after his marriae, !as born a son !ho died  years before his father, at 1/2 his father#s $nal ae." %o! old !as Diophantus !hen he died& ' (et ) * his $nal ae ' a lenthy but simple e+uation ' Final ae * child  youth  bach  marriae  son#s ae   )* ) ) )- ) ' ommon denominator of 127 * 0 )* ) )* ' ) ) ) )- )* ' )* ' ultiply both sides by 0 ) * 045 ' ) * 0 is his $nal ae ' hec' 1  7  12  -  2   * 0 ) 12. he he asoline tan of a car contains -8 liters of asoline and alcohol, the alcohol comprisin 2-9. %o! much of the mi)ture must be dra!n o: and replaced by alcohol so that the tan contain a mi)ture of !hich -89 is alcohol&asoline tan of a car contains -8 liters of asoline and alcohol, the alcohol comprisin 2-9. %o! much of the mi)ture must be dra!n o: and replaced by alcohol so that the tan contain a mi)ture of !hich -89 is alcohol& 1-.i n a chemical laboratory one carboy contains 12 gallons of acids and 18 gallons of water. Another carboy contains 9 gallons of acid and 3 gallons of water. How many gallons must be drawn from each carboy and combined to form a solution that is 7 gallons acid and 7 gallons water? In a chemical laboratory A. one carboy contains 12 gallons of acids and 18 gallons of water. B. Another carboy contains 9 gallons of acid and 3 gallons of water. ----------------How many gallons must be drawn from each carboy and combined to form a solution that is  gallons acid and  gallons water! ---Acid "#uation$ 12A % 9B &  gallons 'ater "#uatio$ 18A % 3B &  gallons -------------------------------------(ol)e by elimination$ *ulti+ly thru 2nd e#uation by 3 to get$ ,A % 9B & 21 ---------------------(ubtract the 1st e#uation from that and sol)e for A$ 2A & 1 A & 13 13 1st carboy gi)es  gals acid and / gals water (ubstitute into 18A % 3B &  and sol)e for B$ 18013 % 3B &  3B & 1 B & 13 13 2nd carboy gi)e 3 gals acid and 1 gal of water &&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 13 of 1st carboy & 1 gals 13 of 2nd carboy &  gals 16. A tobacco dealer mied 12 lbs of one grade of tobacco with 1! lbs of another grade to obtain a blend worth "#$. He then made a second blend worth "61 by miing 8 lbs of the %rst grade and 1# lbs of the second grade. &ind the 'rice 'er lb in each grade. let blend one price be $x Let blend 2 price be y 12x+10y=54 8x+15y=61 2(12x+10y) - 3(8x+15y)= 2*54 - 3*61 24x+20y-24x-45y= -75 -25y=-75 y=3 $3 !or t"e #econ blend pl% t"e &'le o! y in e'tion 12x+10y=24 x=2 $2 $3 i# t"e price 27 (f A)s rate of doing wor* is to +)s is 6,8 and A does a 'iece of wor* in #! days- how long will it ta*e + to do it? /8&,b B& //. 29.  he time re+uired for t!o e)aminees to solve the same problem di:ers by t!o minutes. oether they can solve 32 problems in one hour. %o! lon !ill it tae for the slo!er problem solver to solve the problem& So in 1 min, E ₁ can solve 1/x of a problem.  In one hour, E ₁ solves 60/x problems Similarly, in one hour, E ₂ solves 60/(x − 2 problems !o"e#her #hey solve $ 60/x % 60/(x − 2 & '2 Solvin" for x, an "e##in" x & '/) or *, bu# x & '/) is no# possible since E ₂ solves ('/) − 2 & −1.2* min #o solve a problem. So x = 5  31 A cyclist tra)eled a certain distance at her usual s+eed. If her s+eed had been 2 mihr faster she would ha)e tra)eled the distance in 1 hr less time. If her s+eed had been 2 mihr slower she would ha)e ta4en 2 hr longer. 5ind the distance tra)eled and her usual s+eed. peed__________time__________distance Usual________________r_____________t_____________d Faster______________r+2___________t-1____________d Slower______________r-2___________t+2____________d EQUATIONS - - - Use te second and tird e!uations o" te simpli"ied s#stem and te Elimination $etod to %et &alues "or r and "or t' (ou can ten use tem in an# o" te e!uations to "ind &alue "or d'  (ou can put tis solution on (OU) we*site s#stem o" linear e!uations A c#clist tra&eled a certain distance at er usual speed' I" er speed ad *een 2 mi,r "aster se would a&e tra&eled te distance in 1 r less time' I" er speed ad *een 2 mi,r slower se would a&e ta.en 2 r lon%er' Find te distance tra&eled and er usual speed'  Normal speed: Distance: mph miles  33 . A boat going across a la*e 8 *m wide 'roceeds 2 *m at a certain s'eed and then com'letes the tri' at a s'eed of !.# *mhr faster. +y doing this- the boat arri/es 1! minutes earlier then if the original s'eed had been maintained. &ind the original s'eed of the boat. (et the oriinal speed be ) m / hour  ime taen at oriinal speed * 0 / ) hours  ime taen at increased speed after 2m * 2/ )  6/ 4)8.-5 0/ )  2/)  6/ 4)8.-5 * 18/68 * 1/6 6/ )  6 / 4 )8.-5 * 1/6 ;(<=(> ?@% A ) 4)8.-5 6 364 )8.-5  6 )6 * ) 4 )8.-5 36)  10  36) * )C2  8.-) )C2  8.-) 10 * 8 2)C2  )  36 * 8 2)C2 0) ) 36 * 8 2) 4 )5 4 ) 5 * 8 4 2) 5 4 )5 * 8 ) * m/ hour EABG %BH 0/ * 2 hours 2/ * 8.- hours 6/.- * /3 hours * 1 hour 28 minutes DD * 1 hour -8 minutes Di:erence * 18 minutes 3. coin ban contains I1.6- in nicels, dimes and +uarters. here are t!ice as many nicels as dimes and one more +uarter than nicels. %o! many +uarters are there&  otal Eumber of Eicels * E  otal Eumber of Dimes * D  otal Eumber of Juarters * J Gelations' D*E/2 J*E1 Kalue of Juarters  J  Kalue of Eicels  E  Kalue of Dimes  D * 1.68.2-J  8.8-E  8.1D * 1.68.2-4E15  8.8-E  8.1 4E/25 * 1.6tain t!o as common denominator, !e et, L8.-4E15  8.1E  8.1EM / 2 * 1.68.-E  8.- 8.1E  8.1E * 3.38 8.-E  8.1E  8.1E * 3.38  8.8.7E * 2.08 E* J*E1*1* otal Eumber of Juarters * - . a toy sa/ings ban* contains "17.3! consisting of nic*els- dimes and 0uarters. the number of dimes eceeds twice the number of nic*els by 3 and the number of 0uarters is $ less than # times the number of nic*els. how many 0uarters are in the ban*? ,6%26%3%12,6-1&13 1,6%3 &183 1,6&183-3 1,6&18 1, 6&12 nic4els 1, 26%3 dimes & 2 ,6- & ,/ #uarters ,. ary Gose has I1.- in coins. Ahe has 1 coins in Eicels, Dimes and Juarters.  here are 2 more dimes and +uarters combined. %o! many nicels does she have. <#m oin to assume that "here are 2 more dimes and +uarters combined" means there are 2 more E<HB(A %E dimes and +uarters combined. 1- * -n  18d  2-+ n*2d+ n  d  + * 1 Eo! you have 3 e+uations in 3 variables, !hich you can solve. Aolve the $rst one for +' 1- * -n  18d  2-+ 2-+ * 1-  -n  18d + * 1-/2-  n/-  2d/Drop that + into the second e+uation and solve for d' n*2d+ n * 2  d  1-/2-  n/-  2d/d * n  2  1-/2-  n/-  2d/Drop the d and + into the third e+uation and solve for n' n  d  + * 1 n  n  2  1-/2-  n/-  2d/-  1-/2-  n/-  2d/- * 1 n  n  2 * 1 2n * 16 n*0