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#251 Re: Help Me ! » 6th grade math » 2008-03-13 16:30:03

John E. Franklin got it right.

I completely missed the part about it being the circumference and assumed it was the diameter. My bad.

#252 Re: Help Me ! » 6th grade » 2008-03-13 11:17:39

I like to complicate stuff smile

To be honest, first thing that came to mind was to get common denominators since for some reason I didnt think the fractions would become whole numbers.

#253 Re: Help Me ! » coordinate geometry » 2008-03-13 11:06:26

I did it a little different but you get same result.

First of all I found the distance between A and C to be:

so since this is the diagonal it means that:

Since this is a square then BC and BA have to be of same length so we let u = BC = BA so it becomes

So you know that the length of the sides of the square are sqrt(20). In this way you could find points B and D knowing the distance from A and C is sqrt(20) etc....I guess I overcomplicated this but thats how I got it.

#255 Re: Help Me ! » 6th grade math » 2008-03-13 09:52:14

First of all you need to know what numbers are the prime numbers. For numbers 1-6, the prime numbers are 2,3 and 5. So you have 3 prime numbers with a total outcome of 6 so the probability that the first roll is a prime number is 1/2

Rolling a die is an independent event, which means that rolling the second time doesnt depend on the first roll etc. For numbers 1-6, only 4 and 6 are composite numbers which means that you have a probability of getting a composite number of 2/6 or 1/3.

If you were asking for the probability of first getting a prime number AND THEN getting a composite number on the second roll then just multiply both events together: 1/2 * 1/3 = 1/6

So theres a 1/6 probability of getting a prime number on the first roll and then a composite number on the second roll.

#256 Re: Help Me ! » 6th grade math » 2008-03-13 09:47:11

Find the areas of the two circles and compare them?

You must understand that your dealing with different measurements. Inches and Ft. Convert to one or the other.

40.5 ft = 486 inches.

Thats the diameter of the circle so the radius is half of that, 243.

Area of a circle is equal to: PI times its radius squared.

So the area of the first circle with radius 243in is 185,413.86 inches squared.

The area of the the second circle with radius 80in is 20,096 inches squared.

Clearly the first one has a greater area.

#257 Re: Help Me ! » integration trouble » 2008-03-13 09:00:49

My professor always tells us to change the limits but for me its more work and I never do it. Besides, I have gotten a couple questions on tests wrong because I changed the limits of integration where I couldnt because of sin, cos etc. And to be honest, I really cant tell where I can change the limits or not, so I never change them and plug them in when I have converted the answer back to x (or the initial variable). Works out for me.

#258 Re: Help Me ! » integration trouble » 2008-03-13 08:56:36

Yes. That is called "trigonometric substitution" which is used in integrals that have square roots in it which otherwise would be impossible to do, or way harder.

Part c could have been done with the same trig substitution.

#259 Re: Help Me ! » integration trouble » 2008-03-13 08:39:33

For part d:

Plug this in a calculator and it gives you = 0.159435246248

Does this answer part d or I missed something?

#260 Re: Help Me ! » integration trouble » 2008-03-13 08:12:42

Note: I was trying to write out the whole problem but somehow Latex doesnt seem to work. I must be dumb. This always happens to me. Can anyone point out thats wrong?

#261 Re: Help Me ! » Circles » 2008-03-12 02:04:31

Ok so everything but the square is shaded. So it would be something like this:

60707714jz3.jpg

Ok so what you need to do is find the area of the whole circle which is:

A = PI * radius^2 = 6.1544 cm^2

And then you find the area of the square which is:

A = l * w = 3.61 cm^2

So the area of the part that is shaded is equal to the area of the whole circle minus the area of the square inside (which is not shaded) so:

6.15 - 3.61 = 2.54 cm^2

So 2.54 is the area of the section that is shaded. To find the ratio (percentage) of the section that is shaded in respect to the circle, it would be:

2.54/6.15 = 0.413 = 41.3%

Hope that helps.

#262 Re: Help Me ! » Circles » 2008-03-12 00:52:02

Im not sure what your asking?

Theres a square inscribed inside a circle. Are you saying the square is shaded and you want to know what percentage is shaded?

#263 Re: Help Me ! » How should I integrate this? » 2008-03-11 13:34:49

Thanks Daniel123.

So I guess Gotovina got her answer and more than she asked for....:)

#264 Re: Help Me ! » How should I integrate this? » 2008-03-11 07:09:07

Ok I fixed the first part and tried to do the same to the second but still showing up messed up?

#265 Re: Help Me ! » How should I integrate this? » 2008-03-10 12:32:47

Mathsy, could you tell me whats wrong with my code?

#266 Re: Help Me ! » How should I integrate this? » 2008-03-10 12:01:48

Yes by parts.

The point of integration by parts is letting u equal a term that when you integrate it, it becomes simpler, in this case x.

so by integration by parts this is equal to:

#267 Re: Help Me ! » Solve using algebra » 2008-03-09 23:13:21

This calls for a system of equations.

So this investor has $50,000 to invest in two different ways. The regular one and the more risky.

let x = regular invesment
let y = risky investment.

You know the total of both investments is $50,000 so:

x + y = 50000

You also know how much each investment yields.

0.08x + 0.18y = 5600

That means that the 8% yield on the regular investment plus the 18% yield on the risky investment is equal to $5600. This was given to us.

So now you have 2 equations with x and y. So now just solve both variables.

0.08x + 0.18y = 5600
x + y = 50000

y = 50000 - x

0.08x + 0.18(50000 - x) = 5600
0.08 + 9000 - 0.18x = 5600
-0.1x = -3400
x = 34000

y = 50000 - x
y = 50000 - 34000
y = 16000

So now you know that this investor invested $34000 in the regular investment and $16000 in the more risky investment.

Hope I helped.

#269 Re: Help Me ! » Need help with this mathematical induction problem. » 2008-03-08 10:21:44

Yes that is correct. You should say that {2^(2k+1)+3^(2k+1)} is divisible by the assumption and not from the previous step. Remember that the whole point of induction is to prove that k+1 follows from k.

#271 Re: Help Me ! » Math vs Physics » 2008-03-07 13:06:16

I had this dilemma but with math and computer science. I chose math.

You can always choose one and get a minor in the other. Which is what im doing now.

#272 Re: Help Me ! » I need help With Algebra I just don't get it. » 2008-03-07 13:04:25

As Daniel123 is saying, you just cant come in here and say you dont understand a broad topic. Please ask specific questions so that we can help you.

#273 Re: Help Me ! » How to find this integral? » 2008-03-03 09:59:09

Could anyone help?

Im really confused with this Partial Fractions in Calculus.

#274 Help Me ! » How to find this integral? » 2008-03-03 04:12:35

LuisRodg
Replies: 1

Have to do it using partial fractions so:

I have to give the answers in terms of A,B,C which means I dont have to solve for the constants but actually find the integral. So all this integrals are equal to:

and now im lost?

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and what about this one:

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