no, i dont think anyone wants to do your homework for you

This ain''t my homework. In Pre-Calculus? Yeah, right...

Also, notice how i said i won''t post a new question until the old one was answered? Does that sound like I have a deadline?

I just wanna see who answers it, and how quickly.

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A man with no head is still a man.
A head with no man is plain freaky.

Close, but no cigar.

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A man with no head is still a man.
A head with no man is plain freaky.

Hmmm... so they increased the dosage 20%, and as a result 1/3 of the patients fatally OD''d? Considering this, I''d think the surviving patients would refuse to take any more of this medicine, so the hospital''s remaining supply should last indefinitely.

(sigh)

=======================================
A man with no head is still a man.
A head with no man is plain freaky.

The number of tablets is logically t*d*p: the patients, times the days, times the dosage. So if

(tablets) = (patients) * (days) * (dosage)

then:

(days) = (tablets) / ((days) * (dosage))

(days) = (t*d*p) / ((2p/3) * (6d/5))

(days) = 5d/4 (edit for stupid arithmetic error!)

I think.

-Ironblayde
 Aeon Software

Edited by - Ironblayde on September 4, 2000 4:10:34 AM

IronBlayde - OHHH, so close! Try again!

=======================================
A man with no head is still a man.
A head with no man is plain freaky.

Curses! I can't believe I got the logic right and then made a stupid arithmetic error. That should say 5d/4, right?

-Ironblayde
 Aeon Software

Edited by - Ironblayde on September 4, 2000 4:09:25 AM

5d/4 IS CORRECT!

Next problem :

If a and b are the legs of a right triangle whose hypotenuse is 10 units long and whose area is 20 units-squared, what is the value of (a+b)^2?

=======================================
A man with no head is still a man.
A head with no man is plain freaky.

By the Pythagorean Theorem, and the formula for the area of a triangle,

a² + b² = 100
.5ab = 20

So (a+b)² = a² + b² + 2ab = 100 + 4(20) = 180

?

-Ironblayde
 Aeon Software

I could be way off base (and Ironblayde, you obviously know math), but I don't think that the pythagorean theorem is *needed* in this case. I believe the hypotenuse is given to say that either leg can not be >=10. So now, with A = 1/2bh, we know that the product of the legs has to be 40. The only two (I think!) numbers that can be multiplied to get 40 are 5 and 8.

(5 + 8) ^2
25 + 80 + 64
169?

?

Martin

Edited by - lpsoftware on September 4, 2000 11:13:31 AM

Edited by - lpsoftware on September 4, 2000 11:25:04 AM

Are you sure you mean (a+b)^2 and not a^2+b^2 ? Because (a+b)^2 = (a+b)(a+b) = a^2 + 2ab + b^2, which is not the same thing. So do you mean (a+b)^2 yes?


Ironblayde is right then. Otherwise I can''t seem to solve it.

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#pragma twice

sharewaregames.20m.com

Yep, Ironblayde nailed it again. The answer was 180. Lets see, a HARD one this time...

To what percent of 60*sqr(5) is sqr(3)% of 5*sqr(60) equal?

AND

The circumference of circle II is 2 feet longer than the circumference of circle I. How many feet longer is the radius of circle II than than the radius of circle I?

*Answer in fraction form

=======================================
A man with no head is still a man.
A head with no man is plain freaky.

1.) sqr(3)% of 5sqr(60) is:

5sqr(60) * (sqr(3))/100 = (3sqr(5))/10

which is the same percentage of 60sqr(5) as 3/10 is of 60.

3/10 divided by 60 is .005, which is (1/2)%. (I think )

2.) Let C be the radius of circle I. Then the radius of circle I is C/(2pi), and the radius of circle II is (C+2)/(2pi). Subtracting the former from the latter yields 2/(2pi), or 1/pi feet.

-Ironblayde
 Aeon Software

Edited by - Ironblayde on September 4, 2000 11:03:24 PM

BTW, the edit was for another dumb error I just realized I made.

-Ironblayde
 Aeon Software

I''ve got a math question. It was in my math book, but I''m writing it from memory, so its not exactly the same. And we already discuessed it in class, so its not my homework, and I have the right answer.

You have a cup of coffee and a cup of cream. You take a spoon of the cream and mix it in the coffee. Then you take a spoon from the cup of coffee and put it in the cream. Now is there more coffee in the cream cup or cream in the coffee cup?

BTW, when I said a cup I meant the measurement.

I''d think cream in the coffee, because you put pure cream in the coffee with the spoon, but you put diluted cream/coffee hybrid back in the cream.

That took me too long to answer (two minutes, about).

1.)sqr(3)% of 5*sqr(60)=5*sqr(60)*(sqr(3)/100);
(5*sqr(60)*(sqr(3)/100))/60*sqr(5)=0.005
0.005% is the answer. This is easy to see because (5*sqr(60)*(sqr(3)/100)) is so much smaller than 60*sqr(5).
2.)P=PI;
C_II=2Pr_II;
C_I=2Pr_I;
C_II-2=C_I;
C_II-2=2Pr_I;
2Pr_II-2=2Pr_I;
r_II-1/P=r_I;
radius of circle II is 1/P foot longer.

Coffee.)
You have an equal amount of coffee and cream to begin with. You have a spoon, say a teaspoon, that removes the same amount of liquid from each cup. You take a spoon of pure cream and mix it into the coffee. You now take a spoon of coffee/cream and put it into the cream. Both have the same volume of liquid. There is more cream in the coffee because you put less coffee into the cream than you put cream into coffee. You put less coffee in the cream because it was not all coffee that you put into the cream, some of the volume of the spoon was taken up by some of the cream you mixed into the coffee.

More cream in coffee.

These are the answers. What is the next question? bring it on, I can answer it.

*HERE IS THE EDIT*
I did not look at the answers that were posted before, so don't say I stealed from you.
These questions are simplistic. Here's one for you, what is the derivative of arcsin(x)? Here is a hint, sin(arcsin(x))=x. The derivative is defined as the limit as h approaches 0 in the function g(f(x))=(f(x+h)-f(x))/h;

I have a cannon, and it launches potatoes.
You will bow down to me, or you will eat mashed potatoes at 60mph.
EAT AT WAFFLE HOUSE!
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Edited by - webspynner_99 on September 6, 2000 9:03:57 PM

The answer is not more cream in the coffee cup.