itsyaboi

What's the calculation?

whoops, i just realized :P

I could try, but you need to define p,q,r,s.

are you trying to go for their coordinates?

in that case, I can try barycentric coordinates.

An even perfect square is produced by an even number. The smallest, being 2, which 2^2 = 4.

Then, the largest, being 30, which is 30^2=900.

Counting all of the even numbers from 2-30 yields your answer is A.15. 

Answer:

speed: 5 m/s
height: 9.99 m 

Explanation:

KE=12mv2
kinetic energy = 12 * mass * velocity squared

12.5J=12⋅1kg⋅v2
1⋅v2=25J
v2=25J
v=±5J

since speed does not specify direction as velocity does, we can take the value of 5m/s for speed.

GPE=mgh
gravitational potential energy = mass * gravitational field strength * height

assuming the airplane is flying above Earth, the gravitational field strength is 9.81N/kg.

98J=1kg⋅9.81J⋅h

98J=9.81h

9.81h=98

h=989.81=9.99

h=9.99m 

3/5 = 12/x, x=20.

read this: https://artofproblemsolving.com/wiki/index.php?title=2004_AIME_II_Problems/Problem_4

First, multiply by the conjugate of 1-x, which is $\frac{1-x^2}{1+x}$. 

Then, simplifying things, we get $\lim\limits_{x\rightarrow1}(-\frac{x \ln x}{x-1})$.

Which then turns into this: $-\lim _{x\to \:1}\left(\frac{\frac{1}{-1+x}\ln \left(x\right)}{\frac{1}{x}}\right) \\ \frac{\lim _{x\to \:1}\left(\frac{1}{-1+x}\ln \left(x\right)\right)}{\lim _{x\to \:1}\left(\frac{1}{x}\right)}$

Using L'Hopital's rule and simplifying, we get that this equation is equal to 1. 

We also need to find the value of $\lim _{x\to \:1}\left(\frac{1}{x}\right)$ which is easily equal to 1.

Therefore your answer is $-\frac{1}{1} = \boxed{-1}$.

m must be four, since 4^2 = 16. Then, 2mn = degree 1 term, which means mn = 18.

50 times 25 - 50 - 50 times 5 - 50 times 10 = 50 times 9.

A is your answer.