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Solution of practice problems

# Solution of practice problems - 2300 Final Exam Practice...

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Unformatted text preview: 2300 Final Exam Practice Problems Formulas which are provided on the ﬁnal exam are at; the end of this practice ﬁnal, 8 1. Find the slope of the line containing the points (—2,4) and (6, ——3). A4 (gig—g: 0.; D. —% E. —§ .3; .. qt. , "7 9233 v .._____ Y; ' "I; 6 w ('2) 3’ W1. 2. Suppose 280 tons of corn. were harvested in 5 days and 940 tons in 20 days. between tons T and days d is linear, express T as a function of d. A. TM) = 5d + 280 B. T(d) = —44d + 500 T(d) 2 44d + 60 D. T(d) = 60d + 44 E. T(d) _—. 44d — 60 ' r a a ﬁlm; jaw. 15% “f oi f If the relationship . Tr— Mcl'zi' c Fimn '1. 3. When 30 orange trees are planted per acre each tree yields 150 oranges For each additional D tree per acre, the yield decreases by 3 oranges per tree. Express the total yield of oranges per acre, Y, as a‘function of the number of trees planted per acre, 9:, if :1: 2 30. A. Y=4500+60w—3:1:2 B. Yz§z+80 0. Y1: 15037—3552 @Y2240m—3m2 E. Y2900+3m~60m2 . x u} Total jwhl vf' 17/07“ fry 0!le Tit: ’lWLUéj’h'exfnm’ef X 3%“ ”ﬂ 8““ 3m me [Se ~ auf—wl WWW "”‘"‘”7\‘"aa(,/€l.‘enu.( nwém 7% MM 36W 30 nee; ‘) Y; {C 2417 ﬁx): 2%oX-j3x7' C, 4. A manufacturer can sell dining-room tables for \$70 apiece. The manufacturer’s total cost consists of a ﬁxed overhead of \$8000 plus production costs 0173930 per table. How meny @9135“ g...».w......4 Mi. ' must the manufacturer sell to EEEE‘MEXQP? WVVX A. 80 33. 267 @ 20.0 D. 20 E. 136 . Wen/(1‘ gym meg“; felt-«‘1 0 . 7% it :. yevenewe, w .»c.é3-t Yew-ewe 2:; fry-kc x. rim-66! a}! W Mlle £9111 ‘70 X Mt {\$er Walked '1' cm fey we )1 new é] {4w “6er 5’44 . '" 37“? “l“ gay I ' PM?” 7634f - Was-WWW :9 53> 4*a'vfw‘3’we '55 “memo 5 8- 5. Iff(:c :»\/:Is.+1a11dg(m):;r- +7thcn (f(g(-1))='- A.0 .3 aﬁ D.7 E.¢§+1 3mm)“ 3 NW”) N) W D 6 1mm: :than2 3 2 2 2 2 At—Ema+h_; 0' 95(9:+h) _x(m+h) E"(m+h)2 ' w ' ' 2. ML... . ‘ mvgcmz f'W)’M s“ h w x . w MK”) ‘ I '1 A 7. The domain of ﬂax) 2: 39: 1 is: @w<10rm>1 B.a:>1 0.1:)0D.m<0mrm>0E,——1<a:<1 arr—pine a» Maﬁa :5 mm w W m . m2+4\$—5 3‘} \$0)»; Q a ‘ ‘\ ' ‘ . 8‘93 {52—1 : W w "‘6'“ 4‘3? "920! Sthrlyf»j f‘f’i5( ‘ Am [email protected] 11—3 Es 2 . " I I 1““ . X {3* 5’ 1 ‘NM Mi? 1 ”M 7{ {'5‘ ' H 5‘ my x 4 x~> , (w) W" m t x+1 ” (+1 ‘” 3 9. lim 6“” = 1400 I: @O.B.iC.—1D.00E.e of. \$16 Suppose ~332+2 ifmZO Find all val where the function f is discontinuous. ' 272—1 . B f(:z:)={:6—1 1fm<0 > 2 A. a: 2 —l a: z 0 C. m r: 1 D. a: r: 2 E. f is continuous for all values of m. X?! .91 w ”WC?” é? JBCmﬁﬁww_ at x3,( ﬂaw (s (MM ' (2 ((2 w: §§5€3(% ﬁe. (WWW-U fdmb (a . *EJM wawwm 2.32 gm]; W} %&'%.(.%%%2. (a? . '15“ n. 233, s V - gmw :ﬁg‘ﬁ? Er \$9M gs“) %E% %%a Y :32 Y3?“ (4%.memew 7%“! ” U). R g . ( ” .. mmﬂ‘, Y4 3 W New 35% 2%. {46 «mva ‘4” 4‘7” W MY} 25 Mam 9M Y<v »« . ”1%,. gm “31.1.... g as w E 11. Find all values of a: for which the function f (as) v: 2:133 — 32:2 -— 123: + 12 is increasing. A. -1<m<2 B. :z:< —1 c:.m>[email protected];2m<—1.mu.<1x>2 E.:I:<2a,ndm>2 Hﬁhéx‘ﬁéxvn v :2; 51736.22ng ﬂ. 5124') mm v 3522 my W2”? WNW 355122». 2&2, ”wipe/z [he , f”) 7"? {a may“. i 22 (WW, '4). Mo! (.2, M”) w‘lcw x42»? aw! K221. ' g 2 A 12. The derivativheigkl“. + 1 is: w+5 2:2 +10a: — 1 2:52 + 10:1: 3:152 + 10:1: +1 ‘ .. —-—-— . .-—-—-D.———-— (‘9 (204-5)? B 2” C (m2+1)2 (2+5)2 E. —a;2—10:z:+1 (m4—5)2 22> '222‘2imzimgéc‘2mﬁ? Extmxwia 2 «MW-I ' 2:222 ‘22 (x 2W “ﬂux“; 2| 13. .If y z (3 — 2:2)3 t} m. y” ~- B A. —6m(3 — :22”)2 © 24:);2 '3 —— \$2) — 6(3 — :32)? C5. 6(3 —- .722] D. 24.123 - x2) :4. 12:22 6(3 -— :52) a? mi s n . It, .fj Has-xi“)? (”MU wé s4 WW “g”! was s! MHWW} («255) «~— 6 (*3 “X113?“ ”,1. .m s (s!!!) «é (_ was“ 1 A ' 14. The line tangent to the graph. of f (:L') = a: — —— at a: = 2 has slope: :1! u! 31 {\$213.2 Chg DOE; B 15. Find an equatlma for the tangent; line to the curve: way + my” :1 2 m, the point", (1,1). A. [email protected]+4ym7 Cl. 2m+3y== 5 D. 5512— 211:3 Elm—+3312" \$1614. ﬁg"; ’ a. l 3' Ewﬂu‘l“ aliﬁexmluflm ‘. 7‘ \$22 w!" 1:19 H “l” 7% \$31 ‘19., '9' U 2:: ”v {x ::xy‘)7%< .md’hwm 27W waxy x ”wary“ 0. 2‘2“}(5) «ll ’2, 35 . w mwmawmww’ - Q. ) . . 3 . ‘W‘T H 3 new =33. M "297 - "(MSQQE {m , (39”,) :4"? Hone?) eta-3w»: ”Ml-3?? MMﬂm ‘1 L7) 16. Aftel t yeeus the population of a. certaln town IS P(t)= 50+5t thousand people. A populatiqn . P has an associated CO? level, C(17) 2 (\/PE +1)/2. After 2 yams, the rate at which (3'02 “ "WM mm M... v...“ mw ; 4;; 36m, l. l»! lama! is changing w1th respect to t will be: A s/(af) @130/m C. 56% D 30NWO‘ E. EEO/V3601 WIN! wt villain m3 {Wei ﬁlmy» («MA «mfm h f. -056 - dc if .33me 3 ,. WmﬁlZmW. 7?? .33" t W “WW: 24ml raw .2“ z: m {3:}??? @3133 333 D 17' If 3/132 + ya == :1: — y. Tlmn y’ 2:: A. 1— 2:01; — 33,12 1-3.1 —— 2mg — at? - 3y2 C2. (1- 2wy)/(i’5y2 +1'(1—— 2myj/(m2 + 3112 + 1) E. (23y)/(m2 + 3y2 +- 1) ' WW cliffhwwm; f 7376+ UM) 4w gg‘gﬁ. i“ ﬁx (Mi) «.12 '9 up! 4. .. 'ywzxj H 7% mam 18. If the concentration C(t) of apertain drug remaining in the bloodstream t minutes after it is injected is given by 00:) == t/(Bt2 + 125), then the concentration is a, mggcimgmﬂhcn t = A. 25 B. 15 @5 13.10 E. 20 ﬁt“; Mgfw m) ..... fwﬂﬁf} .. “Hz" '19” \$24.3" 6%) W5? #3:: {,f‘érw wrist" a a m» 5er .‘ 5"”) "<13 9? «W a?” :29" a \$3» M w ‘3'“) {:- y W 2’7 I {VA -{,"{ﬂ LC WM" “; m .. » M ) "ff-{4% » 0-H; ‘35 waﬁmizmi otf fay. ﬁgure m. '5?“ ”a” 5:21 1 W ...
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Solution of practice problems - 2300 Final Exam Practice...

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