Quazars

Ah I see, competely went over my head lol. Still I'm wondering if it is possible solving with Laplace Transforms that's all. How does it apply to squared functions! Thanks

I'm sure I say this every time but you are a life saver. Thank you :)

1% of 67 is one hundredth of that. So you divide by 100 and get 0.67

Now suppose you want 35%, you already have 1% so just multiply by 35.

0.67 * 35 = 23.45

Use the ^ sign. 

For example: 5 to the power of 9 -> 5^9

Per definition of a parallelogram we can see that the two triangles are of exact same size and angles. We also know that the angles of a triangle add up to 180, we can therefore find y by subtracting 33 and 109 from 180.

\(y = 180-109-33=38\)

Because those triangles are similar we can see z and 109 are the same. 

\(z=109\)

We can solve for x the same way by using the similarity principles or by using the fact that a triangle's sum of angles is 180. 

\(x = 180 - z - y = 180 - 109 - 38 = 33\)

(I'm not sure how to edit my post, it's supposed to be 30 instead of 40. Derp)

My reasoning led me to the conclusion that since the last cookie is dependent on the one before it the last one can always be ignored.. to more explain how I mean I'll set up the example with two mathematicians and 10 cookies:

No matter how many cookies x the first one gets(0-10, 11 possibilities) the other one will get 11-x(since all cookies need to be distributed)

- 11 combinations.

When doing this for three mathematicians I follow the same logic that the last persons cookie count is neglectable.

These are the combinations for when the first person gets no cookies whatsoever. (still 11)

But when the first person gets a cookie the amount decreases by one. And for each cookie he(or she) gets the amount decreases...

Continue doing this until the first person gets 10 cookies and the rest none and I ended up with:

1+2+3+4+5+6+7+8+9+10+11. Or \(\sum_{i=1}^{11} i\)

I ended up looking through the case when there are four mathematicians(mostly seeing that depending on the value of cookies the first person had the combination count would be that sum, but 11 would decrease by each cookie the first mathematician had. So I set up the formula you can see in the post. At least that was my reasoning. I don't think N choose K applies here, sorry if I wasn't clear enough in my main post. I also sent you a pm, cheers :)

I think that regardless of base b the n can be multiplied outside the logarithm. So you'd get n*log_b(x). I'm not a 100% sure though.

Kinetic Energy is defined as half of the mass multiplied by the velocity squared, in other words

\(E_{k}= \frac{mv^2}{2}\)

From the question we are given the mass m, which is equal to 47.0 kg. We are also given a velocity v, which is equal to 31.0 m/s. Hence by plugging in the number into the formula we get:

\(E_{k}= \frac{47.0*31.0^2}{2} = 22583.5 \approx 22600 \text{J}\)

Thank you very much CPhil, always a life saver. I always tend to forget that identity >.<

Thank you for this answer. I've done further research and found this pdf which was pretty helpful as well: