Amining company has two mines. one day's operation at mine #1 produces ore that contains 30 metric tons of copper and 600 kilograms of silver, while one day's operation at mine #2 produces ore that contains 40 metric tons of copper and 380 kilograms of silver. let v1 = (30, 600) [vector] and v2 = (40, 380) [vector]. then v1 and v2 represent the "output per day" of mine #1 and mine #2, respectively. a) what physical interpretation can be given to the vector 5v1? b) suppose the company operates mine #1 for x1 days and mine #2 for x2 days. write a vector equation whose solution gives the number of days each mine should operate in order to produce 240 tons of copper and 2824 kilograms of silver. do not solve the equation. c) [m] solve the equation in (b).

a) the output of mine #1 in 5 days

b) x1·v1 +x2·v2 = (240, 2824)

c) x1 = 544/315 ≈ 1.727; x2 = 1482/315 ≈ 4.705

Step-by-step explanation:

a) If v1 represents the production of mine #1 for 1 day, then 5v1 represents that mine's production for 5 days.

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b) The production of each mine, multiplied by the number of days of production, adds together to give the total desired production:

x1·v1 + x2·v2 = (240, 2824)

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c) Treating the vector components separately, the vector equation gives rise to two linear equations:

30x1 +40x2 = 240

600x1 + 380x2 = 2824

These can be solved by any of the usual methods. My favorite for numbers that are large or relatively prime is Cramer's rule and/or a graphing calculator. The above equations can be reduced to standard form to make the numbers slightly more manageable:

3x1 +4x2 = 24

150x1 +95x2 = 706

By Cramer's rule, ...

x1 = (4·706 -95·24)/(4·150 -95·3) = 544/315

x2 = (24·150 -706·3)/315 = 1482/315

mr. graef told us it is 13 pints

step-by-step explanation:

∠acb = 8°

step-by-step explanation:

∠xba lies on the circle and is half the measure of the intercepted arc ax

hence

measure of arc ax = 2 × 42° = 84°

∠acb is outside the circle and is

∠acb = (arcab - arcax ), hence

∠acb = 0.5 (100 - 84) = 0.5 × 16° = 8°