Answer:
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Step-by-step explanation:
Answer:
a. 2,440 km
b. 
Step-by-step explanation:
a. Diameter of Mercury = 
Radius of Mercury = ½ of its diameter = 
Radius of Mercury = 
To convert the radius to ordinary number, multiply 2.44 by 1,000.
Radius of Mercury as an ordinary number = 
b. Mass of Jupiter = 
Mass of Saturn = 
Difference between mass of Jupiter and mass of Saturn =
(subtract 5.68 from 1.898 = 5.68. we would retain the largest exponent, which is raised to the lower of 27.)

Hailey's mixing two different coffee blends. Represent them by x and y (in pounds). Then x + y = 5 lb, and x = 5 - y.How much puree Sum. beans are we talking about here?
0.20x +0.80y = 0.60(5 lb) Mult all 3 terms by 100 to get rid of factions:
20x + 80 y = 300. Substitute 5-y for x:
20(5-y) + 80y = 300 => 100-20y + 80y = 300 => 60y = 200, so y = 20/6 or 10/3 lb den x = 5-10/3, or x =5/3 lb
Use 5/3 lb of the first blend and 10/3 lb of the second blend to come up with 5 lb of a 60% blend.
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:
x = -30
cos(x - 30) = ½
Step-by-step explanation:
Look at the picture