1+2 x 3 - 4 / 5
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@llamonade
Answer: 5.75
Answer: 5.75
x/20+5X5=45
What is x?
3, 2, 1, .....
I knew the answer to that one, but despite being smart at math, I don't wanna answer that @llamonade
For post #8
let a = b
multiply both sides by a
a^2 = a*b
subtract b^2 from both sides
a^2 - b^2 = a*b - b^2
factor both sides
(a + b)*(a - b) = b*(a - b)
divide both sides by (a - b)
a + b = b
replace b with a
b + b = b
2b = b
divide both sides by b
2 = 1
let a = b
multiply both sides by a
a^2 = a*b
subtract b^2 from both sides
a^2 - b^2 = a*b - b^2
factor both sides
(a + b)*(a - b) = b*(a - b)
divide both sides by (a - b)
a + b = b
replace b with a
b + b = b
2b = b
divide both sides by b
2 = 1
Cos90
0
#13
A flywheel is spinning 532 rotations per minute in the counterclockwise direction. A force counteracts this causing it to slow, and eventually turn in the opposite direction (clockwise).
The function describing the wheel's rotations per minute with respect to time is:
2*e^(4t) - 286*e^(2t) - 248
How long must the force be applied before it is spinning at approximately 4.189 radians per second in the clockwise direction?
To convert radians to rotations recall that 1 rotation is 2pi radians.
0
#14
White text hint
Basically just a solve for x type of question in this:
2*e^(4t) - 286*e^(2t) - 248 = 40
I think you learn how to deal with an unknown as an exponent in algebra class (?)
This might require an extra step though. Notice e^(2t) squared is e^(4t).
llamonade wrote:
A flywheel is spinning 532 rotations per minute in the counterclockwise direction. A force counteracts this causing it to slow, and eventually turn in the opposite direction (clockwise).
The function describing the wheel's rotations per minute with respect to time is:
2*e^(4t) - 286*e^(2t) - 248
How long must the force be applied before it is spinning at approximately 4.189 radians per second in the clockwise direction?
To convert radians to rotations recall that 1 rotation is 2pi radians.
4.189/(2 pi) x 60 is Target RPM.
0
#16
Yes 
And since getting an answer out to microseconds probably isn't of any practical use, after you convert it to RPM you can just round to the nearest whole number.
That was my intention anyway.
2*e^(4t) - 286*e^(2t) - 248 = Target RPM
divide all by 2 and set x to e^(2t). Solve for x.
0.5 ln(x) should do it for x>0.
0
#19
Very good!
Letting e^(2t) be a variable, then just solving the normal way with factoring and such was the "extra step" I wasn't sure everyone would know about.
So x = e^(2t) and after factoring x = 144
So now e^(2t) = 144, one equation one unknown, and you can solve for t.
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