Step-by-step explanation:
If she is able to achieve her aim, then the robot is very efficient, as it uses as low as 49 joules in filtering 343 litres of air and water.
It will now be a question of what time the robot is able to achieve this feat, if it is able to do it in a time that matches the quality of the work it does, then it will be reasonable to call the robot efficient, and fast too.
Let L = lengthLet W = width Length is 5 yd less than twice the width ==> L = 2*W - 5 Area is 52 yd^2 ==> 52 = L*W Substitute the value for L from the 1st equation into the 2nd equation
52 = (2*W - 5) * W52 = 2*W^2 - 5*W0 = 2*W^2 - 5*W - 52 0 = (2W - 13)*(W + 4) 2W - 13 = 0 or W + 4 = 0W = 13/2 or W = -4 Width cannot be negative so discard W = -4 W = 13/2 ==> L = 2*(13/2) - 5 = 13 - 5 = 8 So the width is 13/2 yards and the length is 8 yards Check: 2*(13/2) - 5 = 13 - 5 = 8 (13/2)*8 = 4*13 = 52 yd^2
The answer is 7 hope this helps
You multiply 1 1/2 by 5 (because you give it to them 5 times a day) and you get 7 1/2 then you have to multiply that by 4 (because you’re calculating how much over 4 days) and you get 30oz .
