Homework Section

[Daco]

I took the time as 0 since we can assume that at the very beginning of the reaction (time when the solute was added to the water) there is still the maximum (i.e. all) of the solute present in the non-dissolved state. So, 0,09 * 0 is obviously 0 and e⁰ is 1.

BUT BRO YOU DIDINT NOTICE 3' SO THE EQUATION IS BASICALLY 60/5e^0.006t

[duffman]

Drop a soluble product on a glass of water. In instants in (t), in minutes, the amount of the product not dissolved its, in grams, given by the function:



Question n2 - How many time it takes to dissolve 20 grams of the product?

[Daco]

Drop a soluble product on a glass of water. In instants in (t), in minutes, the amount of the product not dissolved its, in grams, given by the function:



Question n2 - How many time it takes to dissolve 20 grams of the product?

i don't understand the derivative of 3 is 0 why bother putting it again?????

[duffman]

i don't understand the derivative of 3 is 0 why bother putting it again?????

fuck my teacher man, dam hoe

[Daco]

wait if 3 doesn't exist then the second question is wrong

[Patton]

BUT BRO YOU DIDINT NOTICE 3' SO THE EQUATION IS BASICALLY 60/5e^0.006t

[Daco]

shit the whole thing is wrong

[duffman]

wait if 3 doesn't exist then the second question is wrong

Probably you should use a graphic calculator, its something about with LOG() / Y= / TABLE... But I don't remember now..
shit the whole thing is wrong
[/quote]
YOU GUYS SUCK MAN

[Daco]

Probably you should use a graphic calculator, its something about with LOG() / Y= / TABLE... But I don't remember now..
shit the whole thing is wrong

YOU GUYS SUCK MAN

yes but the question doesn't make sense given that 3' = 0 (supposing there is ' even) the initial product is 12 grams and the question is about dissolving 20grams which isn't logical. at all
before you make any logs or anything

[ElMartu]

daco get on msn or something man

[Daco]

if there isn't '

initial product is q(0) = 60/(5-3) = 30grams
dissolve 20 grams means q(t)=30-20=60/(5e^0.09t-3) => 5e^0.09t-3=6 => 5e^0.09t = 9 => e^0.09t = 9/5 => 0.09t = ln(9/5) => t=ln(9/5)/0.09 = 6.53096294 minute

FUCK OFF

[duffman]

if there isn't '

initial product is q(0) = 60/(5-3) = 30grams

True

Question n2 - How many time it takes to dissolve 20 grams of the product?

y1=60/5e(0.09t)+3
y2=20

^On the calculator and you will see the graphic and then you will see the intersetion and then the result - 2,202.

[Daco]

go suck a duck man, go learn maths

3 has to be there else the whole question wouldn't mean sense. Say we take consideration of the apostrophe on top of the 3. The equation of the remaining product becomes:

q(t)=60/(5e^0.09t)

Initial quantity: q(t=0)=60/5=12g

Question 2: How many time it takes to dissolve 20 grams of the product?

When you read the question you see it doesn't mean any sense since the initial quantity is 12g and he's asking for the amount of time to dissolve 20g. If you go ahead and calculate: e^0.09t = -60/40 (doesn't mean sense) ; t=ln(-60/40)/0.09 (impossible)

[Patton]

Question 2: How many time it takes to dissolve 20 grams of the product?

When you read the question you see it doesn't mean any sense since the initial quantity is 12g and he's asking for the amount of time to dissolve 20g. If you go ahead and calculate: e^0.09t = -60/40 (doesn't mean sense) ; t=ln(-60/40)/0.09 (impossible)

It makes sense since the solution can behave according to the function no matter what amount of solute is added into the solvent; it doesn't have to relate to the first question although the result from the first question can come in handy when checking the result you get. The problem is that the function doesn't allow you to determine how much time it takes to dissolve something as it shows the amount of remaining solute. You could try solving with q as 0 since then all solute is dissolved, but then you wouldn't be able to solve the equation. :razz:

[Daco]

It makes sense since the solution can behave according to the function no matter what amount of solute is added into the solvent; it doesn't have to relate to the first question although the result from the first question can come in handy when checking the result you get. The problem is that the function doesn't allow you to determine how much time it takes to dissolve something as it shows the amount of remaining solute. You could try solving with q as 0 since then all solute is dissolved, but then you wouldn't be able to solve the equation. :razz:

t = +oo

checkmate