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Hello young person!. That is very simple. Since Pu-238 has a half-life of 87.7 years, you have to figure out how much Pu-238 has decayed, through radioactivity, in that 19-year period. I hope you understand what half-life means. It simply means that if you had 100 grams of this Pu-238, in 87.7 years, there will be only half of it left , or just 50 grams. So, we try to find out how much has decayed and how much is left over. So, after 19 years, we have 19/87.7=0.216647 or21.66% of 1 half-life gone in 19 years. So, 1 half-life - 2^(-0.216647) (the portion that has decayed) =.86063 or 86.063% remains.

So, if you have 2500 grams, after 19 years only: 2500 x .86063=~2,151 will have remained. The formula used to do this is: A(r)=A(o) x 2^(-n/half-life), where A(r)=amount remaining after n years, A(o)=original amount of material you started with, n=number of years elapsed.

They should have taught this in school when you are dealing with radioactive elements such as Pu-238. I hope you understand it a little better now.

Here's (a)

Setting 

Ps = 4 Q + 100,

Pd = 400 − 2 Q     equal, we have

4Q + 100  = 400 - 2Q     add 2Q to both sides, subtract 100 from both sides

6Q  = 300     divide by 6 on both sides

Q = 50   this is the equilibrium quantity

And using either equation, the equilibrium price =  4(50) + 100  =  200 + 100  =  300  [euros]

When multiplying expressions with exponents that have the same base (which would be 4 in this case), just add the exponents and keep the base the same.

6+(-8)=-2, so (4^6)*(4^-8)=4^-2

In binary form  2 is written as 10.     So   2  (written as 10 in binary )   minus 2  = 0        

csc (1.6)   first quad......use the csc inverse....so we have   acsc(1/6)  = θ = 38.682187453489°   

csc (1.7) second quadrant   ........   =  180 - acsc(1.7)  = θ = [180 - 36.031879072471]°  = 143.968120927529°

The csc cannot be positive in the 3rd and 4th quadrants.......thus......the last two have no solutions

Thank you so much for everybodys help! The fifth response is really the only one I can understand, but where on earth, for the first question, do you pull out a 1-2^-(19/87.7) ?? Where does that equation come from?

But what is n?

it doesn't  say she actually gives the cookies so she loses none

4^-2

Sorry, I made a math error in part c):

c) in this scenario, the 2500 grams would decay to 1 - 2 ^(-25/87.7) =82.07% x 2,500 =2,052 grams.

This would only generate: 2,052 x .56 =1,149 watts of thermal power. But to generate 85 watts of electrical power, the engine would have to have an efficiency of :85/1,149 =7.4%. Now, all 3 of us get the same figures. Any minor difference you see are due to rounding off of certain numbers.

"The closed form sum of 12[1^2*2+2^2*3+...+n^2(n+1)] for n>=1 is n(n+1)(n+2)(an+b). Find a and b."

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I was about to comment on your result for part c, but you corrected it before I managed to make the comment!

NASA is designing a robotic spacecraft to visit Neptune. The probe will use the radioactive isotope Plutonium-238(Pu-238) as a power source. One gram of Pu-238 generates 0.56 watts of thermal power and has a half-life of 87.7 years. The generator that converts thermal power to electrical power for the probe's instruments operates at 6% efficiency (i.e. 1 watt of thermal power generates 0.06 watts of electrical power). The instruments on the spacecraft require 85 watts of electrical power and need to operate for 19 years to complete the mission. 
a. How many whole grams of Pu-238 are required to complete the mission? 
b. Discuss how a different generator, operating at 7% efficiency, would affect the amount of Pu-238 required to complete the mission. Use algebra to back up your claim. 
c. Suppose NASA wants the mission to last 25 years and has only 2500 grams of Pu-238 available for the probe. To what operating efficiency must the engineers design the generator? Round your answer to the nearest tenth of a percent.

OK, young person, I'm no "rocket scientist", but will try to help you as much as I can:

a) Since the generator operates at 6% efficiency, that means in order for the generator to generate a continuous power of 85 watts, it will require:

85/.06=1,417 watts of thermal power. But, we know that one gram of Pu-238 generates 0.56 watts of thermal power and has a half-life of 87.7 years. We, there require:1,417 / .56=2,530 grams of Pu-238. But the mission will need to last for 19 years. So, the 2,530 grams will decay in that period to:

1 - 2^-(19/87.7) =86% x 2,530, which will reduce the 2,530 grams of Pu-238 to about 2,177 grams, which may compromise the 19-year mission, especially towards the end when it is transmitting all its data back to Earth. Therefore, in my humble opinion, it would be prudent for them to plan for this known problem and add an additional Pu-238 by the above percentage. So, the theoritical quantity of 2,530 grams should be increased by:2,530 / .86 =2,942 grams, just to be safe.

b) If the generator efficiency were at 7%, it would reduce the amount of Pu-238 by about 16%. They would require:85/.07/.56 =2,168 grams would be needed in that case plus allowing for the decay:2,168/.86=2,521 grams of Pu-238 would be needed.

c) in this scenario, the 2500 grams would decay to 1 - 2 ^(-25/87.7) =71.5% x 2,500 =1,787 grams.

This would only generate: 1,787 x .56 =~1,000 watts of thermal power. But to generate 85 watts of electrical power, the engine would have to have an efficiency of :85/1,000 =8.5%. I think, that is as far as I can take it. Good luck to you. Hope it helps.

c. Suppose NASA wants the mission to last 25 years and has only 2500 grams of Pu-238 available for the probe. To what operating efficiency must the engineers design the generator? Round your answer to the nearest tenth of a percent.

Rinese doen't give away any cookies.

Here's the answer to part a.

This confirms Omi67's result.

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I started doing 0.56*.06*x=85 for the first question. Is this right? Do I need to do anything with the 19 years thats mentioned in the problem? What about b. And c. ? This problem is so confusing.

Thanks, Omi67.......I didn't take the question this far since they only wanted to know how many values there were......I forgot that we could use the zeroes to find the exact polynomial....good thinking...!!!!

The total arrangements possible = 5!  = 120

If the two girls sit at both ends, we can seat the girls in two ways and everyone else in the middle 3! = 6 ways. So, the number of possible arrangements =  2 * 3!  =  12

And we can choose any 2 of the 3 boys to sit at both ends and we can arrange them in two ways. And for each of these arrangements, everyone else in the middle can be arranged in 3!  = 6 ways. So the number of possible arrangements =  C(3,2) *2 * 3! *   = 36

So........the probability that a girll sits at both ends or a boy sits at both ends = [ 12 + 36] / 120  = 48/120   = 2.5  = 0.40

The graph of y=f(x) is shown in red below. For how many values of c is f(c) = -1? (Assum grid lines are spaced 1 unit apart)

We can construct a square with its  bottom left edge at (0,0), its top left edge at (0,2), its top right edge at (2,2)  and its bottom right edge at (2,0)....all the points within this square will have x, y coordinates between 0 and 2

So......the total area of this square  = 4 units^2

And the probability that a point in this square lies exactly 1  unit from the origin will be the collection of points in the quarter circle contained in this square with a radius of 1 unit......and the area of this circle = (1/4)pi (1)^2   = pi/4

So....the probability that any chosen point lies less than 1 unit form the origin will be a little less than :

[ pi/ 4] / 4  = pi / 16  ≈  .196

Here's a picture of the situation :

Hey, rarinstraw........!!!!!........missed you.....and glad you're back.....!!!

EV=logbase2(f^2/t)

EV =  log₂ f^2  -  log₂ t

EV = 2 log₂f  - log₂ t

EV = 2 log₂ (5.657 ) - log₂ (1/60)  ≈ 10.90696

So

10.90696  = 2log₂ (f)   - log₂ (4/60)

10.90696 + log₂(4/60) = 2 log₂ f         divide by 2

[ 10.90696 + log₂(4/60) ]/ 2   = log₂ f

This says that f  =  2^ ([ 10.90696 + log₂(4/60) ]/ 2)  ≈  11.3140

[ I hope this is a valid f stop !!!  ]

Remeber the rule that      a^m / a^n =    a^(m - n)   ......so we have

2^(x+2)/2^2

2^(x + 2  - 2)   =

2^x