Width: x
Length: 2x
Area=x(2x)=2x^2
2x^2=200
x^2=100
x=10 or -10(reject as x>0) [width cannot be less than 0, doesnt make sense]
Perimeter
=x+x+2x+2x
=6x
=6(10)
=60 yd
Answer:
The first box will have 90 marbles, the second 180 and the third 270.
Step-by-step explanation:
Division in a ratio of 1:2:3
1 + 2 + 3 = 6
So
The first box will have 1/6 of the marbles.
The second box will have 2/6 = 1/3 of the marbles
The third box will have 3/6 = 1/2 of the marbles.
First box:
One sixth, so:
(1/6)*540 = 540/6 = 90
Second box:
One third, so:
(1/3)*540 = 540/3 = 180
Third box:
One half, so:
(1/2)*540 = 540/2 = 270
The first box will have 90 marbles, the second 180 and the third 270.
Let S(t) denote the amount of sugar in the tank at time t. Sugar flows in at a rate of
(0.04 kg/L) * (2 L/min) = 0.08 kg/min = 8/100 kg/min
and flows out at a rate of
(S(t)/1600 kg/L) * (2 L/min) = S(t)/800 kg/min
Then the net flow rate is governed by the differential equation
Solve for S(t):
The left side is the derivative of a product:
![\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/800}S(t)\right]=\dfrac8{100}e^{t/800}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dt%7D%5Cleft%5Be%5E%7Bt%2F800%7DS%28t%29%5Cright%5D%3D%5Cdfrac8%7B100%7De%5E%7Bt%2F800%7D)
Integrate both sides:
There's no sugar in the water at the start, so (a) S(0) = 0, which gives
and so (b) the amount of sugar in the tank at time t is
As 
, the exponential term vanishes and (c) the tank will eventually contain 64 kg of sugar.
Answer:
The equation i came up with is $55= 7.8x - 47.20
