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Question Number 98306 by me2love2math last updated on 12/Jun/20

Commented by me2love2math last updated on 12/Jun/20

precalculus....q1 and 2 pls

$${precalculus}....{q}\mathrm{1}\:{and}\:\mathrm{2}\:{pls} \\ $$

Answered by Rio Michael last updated on 12/Jun/20

f(t) = −0.005(t−12)(t+3)(t+0.85), 0 ≤ t ≤ 28 (a) f(7) = −0.005(7−12)(7 + 13)(7 + 0.85) = 11.35 grams/hour Do the rest fof t = 13 h, t = 17 hours, t = 26 hours. (b) when the growth rate is 0 , the f(t) = 0 ⇔ −0.005(t −12)(t +3)(t + 0.85) = 0 ⇒ t = 12 h at t = 0, f(0) = −0.005(−12)(3)(0.85) = 0.05 (d) f ′(t) > 0 and f ′(t) < 0

$${f}\left({t}\right)\:=\:−\mathrm{0}.\mathrm{005}\left({t}−\mathrm{12}\right)\left({t}+\mathrm{3}\right)\left({t}+\mathrm{0}.\mathrm{85}\right),\:\mathrm{0}\:\leqslant\:{t}\:\leqslant\:\mathrm{28} \\ $$$$\left(\mathrm{a}\right)\:{f}\left(\mathrm{7}\right)\:=\:−\mathrm{0}.\mathrm{005}\left(\mathrm{7}−\mathrm{12}\right)\left(\mathrm{7}\:+\:\mathrm{13}\right)\left(\mathrm{7}\:+\:\mathrm{0}.\mathrm{85}\right)\:=\:\mathrm{11}.\mathrm{35}\:\mathrm{grams}/\mathrm{hour} \\ $$$$\mathcal{D}\mathrm{o}\:\mathrm{the}\:\mathrm{rest}\:\mathrm{fof}\:{t}\:=\:\mathrm{13}\:\mathrm{h},\:{t}\:=\:\mathrm{17}\:\mathrm{hours},\:{t}\:=\:\mathrm{26}\:\mathrm{hours}. \\ $$$$\left(\mathrm{b}\right)\:\mathrm{when}\:\mathrm{the}\:\mathrm{growth}\:\mathrm{rate}\:\mathrm{is}\:\mathrm{0}\:,\:\mathrm{the}\:{f}\left({t}\right)\:=\:\mathrm{0} \\ $$$$\Leftrightarrow\:−\mathrm{0}.\mathrm{005}\left({t}\:−\mathrm{12}\right)\left({t}\:+\mathrm{3}\right)\left({t}\:+\:\mathrm{0}.\mathrm{85}\right)\:=\:\mathrm{0}\:\Rightarrow\:{t}\:=\:\mathrm{12}\:\mathrm{h} \\ $$$$\mathrm{at}\:{t}\:=\:\mathrm{0},\:{f}\left(\mathrm{0}\right)\:=\:−\mathrm{0}.\mathrm{005}\left(−\mathrm{12}\right)\left(\mathrm{3}\right)\left(\mathrm{0}.\mathrm{85}\right)\:=\:\mathrm{0}.\mathrm{05} \\ $$$$\left(\mathrm{d}\right)\:{f}\:'\left({t}\right)\:>\:\mathrm{0}\:\mathrm{and}\:{f}\:'\left({t}\right)\:<\:\mathrm{0} \\ $$

Answered by Rio Michael last updated on 12/Jun/20

(2) grams from each substance = Σtotal n^o of grams −Σrequired for each sub

$$\left(\mathrm{2}\right)\:\mathrm{grams}\:\mathrm{from}\:\mathrm{each}\:\mathrm{substance}\:=\:\Sigma\mathrm{total}\:\mathrm{n}^{\mathrm{o}} \:\mathrm{of}\:\mathrm{grams}\:−\Sigma\mathrm{required}\:\mathrm{for}\:\mathrm{each}\:\mathrm{sub} \\ $$$$ \\ $$

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