X/2-y/3=3/2
(6×x/2)-(6×y/3)=6×3/2
3x-2y=9______(1)
x/3+y/2=16/3
(6×x/3)+(6×y/2)=6×16/3
2x+3y=32_____(2)
(1)×3____9x-6y=27____(3)
(2)×2____4x+6y=64____(4)
(3)+(4)___13x=91
x=7
3(7)-2y=9
-2y=-12
y=6
Answer:
what are we trying to do with the problem tho?
Step-by-step explanation:
A few ways. you can enter it into a calculator for one, but the easiest way would be to reference a unit circle and look for an ordered pair where sin (the y value), is equal to -1/2
on a unit circle, the sin value is -1/2 at 7pi/6 and 11pi/6
because sin is the y value, it will additionally be in the quadrants III or IV based on the fact that (-1/2) as a sin value IS a negative and would have to be found in a quadrant in which sin is negative (the lower half of a coordinate plane, in this case)
Answer:
40th and 80th visit
the first visit where you receive both is the 40th
Step-by-step explanation:
personally, I believe that the simplest way to solve this is use an excel spreadsheet to see which visits would include both a free beverage and a free appetizer:
- both rewards would happen only twice, at the 40th and 80th visit.
you can also calculate it mathematically since multiples of 8 and 10 do not repeat themselves very much, only when 8 is multiplied by 5 and 10.
