Ooflord

Let's make two variables, y(your money) and b(brother's money).

The first sentence creates the equation y-5 = b+5

Rearranging this equation gives me y=b+10.

The second sentence creates the equation y+25 = 2(b-25).

I can plug in y from the first equation, giving me (b+10)+25=2(b-25), which simplifying gives b+35=2b-50, which b=85, so your brother has 85 dollars. 

We can check by plugging b in, getting the y value, then plugging it into the other equation. 

y-5=85+5 so y=95

Plugging y=95 and b=85 into the second equation, I get 95+25=2(85-25) which is 120=120, which is true, meaning that I solved everything correct.

So your brother currently has 85. 

First, I simplify \(\frac{ {\left( \frac{1}{2} \right)}^2 + {\left( \frac{1}{3} \right)}^2 }{ {\left( \frac{1}{4} \right)}^2 + {\left( \frac{1}{5} \right)}^2} = \frac{13x}{41y} \) so that it looks like \(\frac{ {\left(\frac{1}{25} \right)} + {\left( \frac{1}{9} \right)} }{ {\left( \frac{1}{16} \right)} + {\left( \frac{1}{25} \right)}} = \frac{13x}{41y} \), and then I added the fractions on the numerator and the fractions on the denominator so I got \((\frac{13}{36})/(\frac{41}{400})\), which can then be simplified to \(13*100/41*9\).

remember that this is equal to \(13x/41y\), so next, I get rid of 13/41 on both sides, giving me \(x/y=100/9\). I want \(\sqrt{x}/\sqrt{y}\), so I square root both sides of \(x/y=100/9\), which gets me\(\sqrt{x}/\sqrt{y}=10/3\).

I'm not really good at expressing things with the LaTeX thing, so I apologize if my answer is unclear, I can try it if you want.

The first thing that I did was simplify ((1/2)^2+(1/3)^2)/((1/4)^2+(1/5)^2), which is (13/36)/(41/400). This can be simplified so that it looks like this: (13/41)(100/9). x/y=100/9, so \(\sqrt{x}/\sqrt{y}=10/3\)

:< It was part of a problem, but I guess I did the first part wrong lol

You can move 3pi/2 and 9pi/13 to one side, and add them together, and then you can cross multiply, it's basic algebra.

Guest is wrong.

The steps to solving it is like this:

We take random sets of three consecutive integers, add them, and find the factors of them.

First set: 1+2+3=6

Second set: 2+3+4=9

Third Set: 3+4+5=12

We can clearly see that the common factor for every set of three is 3.

I hope this helps.

Sorry I had to go. It's not possible to find k, like OrangeJuicy said.

What do you want to find?

Oops, my mistake, sorry D:

Okay, they tell us that f(x) is 3, so lets just bring that into f(x)=x^2 -2x

3=x^2 -2x

We can solve by guess and check, but we can also use the quadratic formula.

Let's move 3 to the other side.

x^2 -2x -3=0

We get 3 and (-1), using the quadratic formula. If you don't know what that is, you can find how to use it on google. I hope you find this useful.

Pretty sure this problem has already been posted... You should find it once you search it up online, it was solved by CPhill.

I should have said inclusive, sorry about that. I guess it depends on whether the problem says inclusive or not. If this problem did, then it includes -2 and 3. Sorry about the mistake I made.

Wait what are you trying to find? The integers between -2 and 3 are just -2, -1, 0, 1, 2, 3. Hope this is what you are looking for.

This is just ratios.

New Age/Classical = 3/1

Classical is 3, from the text

New Age/3 = 3/1

Cross multiply

New Age = 9

Seems like fun, I'll do this later in the day! :D