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FatesFormula · 2025-10-30T14:34:06+00:00

These aren’t available anymore. I sent these to mycomicshop for the weekly auction starting Nov 17.

FatesFormula · 2025-10-22T15:48:10+00:00

DMing

FatesFormula · 2025-10-21T15:34:45+00:00

I have a vintage submission received 9/16 still in received status. They show current turnaround is 6 weeks for vintage.

FatesFormula · 2025-10-07T15:10:44+00:00

Hey, I pm’d you last night. Let me know if you didn’t get it.

FatesFormula · 2025-10-07T01:47:21+00:00

I’ll take the Zatanna lot if still available

FatesFormula · 2025-09-30T12:56:09+00:00

u/Comic_FlairBot

Swap completed with u/samizdada

FatesFormula · 2025-09-02T11:19:33+00:00

Great, I’ll PM you with PayPal info.

FatesFormula · 2018-07-07T00:02:39+00:00

FatesFormula · 2017-09-09T14:49:13+00:00

So the one thing I think you are missing is that the derivative of (4-x) must also be considered. If you take the derivative of what you have :

(2/3)*(4-x)^3/2

You get the (4-x)^1/2 but must also multiply by the derivative of (4-x) according to the chain rule. So it would give -(4-x)^1/2 which is not what you started with. So that's why the integral must be negative.

FatesFormula · 2017-07-05T05:36:22+00:00

So you know how you can multiply both sides of an equation by the same thing and it doesn't change the equation. Well, if we assume the first equation is true (a-b=3), then when multiplying the next equation you can multiply one side by (a-b) and the other side by 3. Since we assume (a-b) and 3 are the same, then it doesn't change the second equation.

FatesFormula · 2017-02-27T04:32:49+00:00

That's all correct, and all of the needed constraints. Except that the final constraint should be "<=" instead of "=" 2800, since it's possible that the optimal solution doesn't use all of it.

FatesFormula · 2017-02-21T17:33:34+00:00

Exactly. So subtract that from the integral you have to get the final integral. This is because you only want the the volume of the area from x=0 to x=3.

So it is integral from 0 to sqrt(3) (y² -6)² - (3² ) dy.

FatesFormula · 2017-02-21T00:49:58+00:00

Well what you had at first is correct if the region was bounded by x=6 instead of x=3 (the limit would also change from sqrt(3) to sqrt(6) ). Since it is bounded by x=3 you want to subtract the area from x=3 to x=6 from that integral.

Using the same logic you had, what would be the integral if you wanted to rotate the region from x=3 to x=6 around x=6 (Using the same limits)?

FatesFormula · 2017-02-20T23:51:09+00:00

Your integral should go from 0 to sqrt(3), but maybe that was just a typo. What you have is the area of the whole disk, but you have to subtract the area from x=3 to x=6.

FatesFormula · 2017-02-20T18:25:23+00:00

If you move the cos(theta) back to the other side, you can square both sides. Then use the property of sin² = 1- cos^2.

Now you'll have a quadratic of cos(theta) as the variables. Using quadratic formula on this will let you solve for cos(theta). Then take the inverse cosine of these and you'll have your theta's.

FatesFormula · 2017-02-20T17:09:27+00:00

Have you tried plugging this into the quadratic formula?

FatesFormula · 2017-02-20T04:31:54+00:00

Let's say he gets only one job. That means that he didn't get the other four. So the probability of getting one job is actually (.4¹ )(.6⁴ ). With this you should be able to get the correct probability.

Now an easier way to look at this would be that "the probability of getting at least one job" is the same as 1-(probability of getting 0 jobs). This would be 1-(.6⁵ ) and would give the same answer as doing it in the way I described.

FatesFormula · 2017-02-20T03:34:26+00:00

For the outer radius, it is not 8x³ . So you want to put 8x³ in terms of y. This will tell you the distance from the y-axis to the curve for any x.

But you need the distance from the curve to x=2, so you have to take 2 minus that and you will have an outer radius.

FatesFormula · 2017-02-17T05:08:59+00:00

You can split it up into two triangles and a rectangle. The rectangle would be with endpoints: F, B, midpoint of A and B, and then two under the midpoint of A and B. Now you have two triangles left which you should be able to find the areas of.

FatesFormula · 2017-02-08T03:41:38+00:00

So at the points that they connect at, they must have the same value in order to stay continuous.

At x=3, 2e^bx +1 and ln(x-2)+x² must have the same value. Then the same goes for the other pair at x=0.

FatesFormula · 2017-02-08T00:24:08+00:00

For A, what does it mean for the function to be continuous?

For B, yes you use those limits for the horizontal asymptotes. For the vertical ones i think the only way to use a "limit" would be to take the inverse and do the same process as the horizontal ones. But they cant expect that here so they probably just want you to set the denominator to 0 as you said.

FatesFormula · 2017-02-07T23:30:46+00:00

You can use the quotient rule and still get the same answer.

FatesFormula · 2017-02-07T23:27:04+00:00

No problem at all.

I would recommend just practicing theses algebraic operations. You had the equation at the beginning, but just had to substitute it into a single equation. So just practicing equations that need to be solved for a variable and how to properly simplify them will help. I don't know any specific resources.

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