RiemannIntegralzzz

You could either a) expand equation 1, b) complete the square on equation 2, c) set up equation 1 equals equation 2 and cancel/simplify

I'll choose completing the square, because expanding is pretty self explanatory.

Completing the square on equation two:

First, factor an 0.6 out of the first two terms, yielding $0.6(x^2+8x)+12$

Second, complete the square on the first two terms by adding a constant, yielding $0.6(x^2+8x+16)-0.6*16+12$.

Simplifying gives $0.6(x+4)^2-0.6*16+12$.

Simplifying further gives $0.6(x+4)^2+2.4$.

Now, setting them equal, we have $0.6(x+4)^2+2.4=0.6(x+4)^2+2.4$, which is true. So, these two functions are equivalent.

Using the formula for area of a rectangle, and converting all mixed numbers into improper numbers, we get $\frac{13}{2}*x=\frac{59}{2}$.

Multiplying both sides by 2 gives $13x=59$.

Dividing by 13 gives $\boxed{x=\frac{59}{13}}$. Or, as a mixed number, $x=4$ $\frac{7}{13}$

Sorry @Melody and @Jamesaiden, they were both wrong. The correct answer was 20. Here was the solution they gave:

The residue of 4n(mod 6) is determined by the residue of n (mod 6). We can build a table showing the possibilities(it doesn't render, but here it is with words)

n (mod 6)             0       1        2           3          4          5                                                          

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4n (mod 6)           0        4       2           0          4          2                                               

As the table shows, 4n=2 (mod 6) is true when n=2 or n=5 (mod 6). Otherwise, it's false.

So, our problem is to count all n between 0 and 60 that leave a remainder of 2 or 5 (mod 6). These integers are 2, 5, 8, 11, 14, 17, ... , 56, 59.

There are $\boxed{20}$ integers in this list.

Thank you both for trying, however! :)

Using Pascal's identity we have $\binom{17}{8}+\binom{17}{9}=\binom{18}{9}=24310+24310=48620$.

Now, to get from $\binom{18}{9}$ to $\binom{19}{9}$, you multiply by $\frac{19}{10}$. To see this, let's write out the binomial coefficients using factorials.

This gives you $\binom{18}{9} =\frac{18!}{9!*(18-9)!}=\frac{18!}{9!*9!}$.

Now, $\binom{19}{9}=\frac{19!=18!*19}{9!*(19-9)!=9!*9!*10}$. Hopefully you should see it by now.

So $\binom{19}{9}=\frac{48620*19}{10}=\boxed{92378}$.

However, there might be an easier way using more identities, so if anybody else has a solution that is easier, please speak up! :)

I think there might be more identites because my solution did not use the fact that $\binom{17}{7}=19448$

Sorry, it was incorrect. I have one more try however.

If the denominator of the circle is $12$, the radius is $6$.

If the center is at $(-6,-3)$, then the equation is $(x-(-6))^2+(y-(-3))^2=r^2$ for some radius r.

Putting these together, we get: $(x-(-6))^2+(y-(-3))^2=6^2$, so $\boxed{(x+6)^2+(y+3)^2=36}$

Putting it in non-standard form by expanding gives $x^2+12x+6^2+y^2+6y+3^2=36$, so $x^2+12x+y^2+6y=-9$. However, the actual equation for a circle is the standard one that was shown first, it is MUCH more common. But if you graph both equations, they give the same circle.

Sorry, it is not rendering for some reason.

Here is the question without Latex:

How many integers n satisfy 0 is less than n is less than 60 and 4n=2 (mod 6)

Ohhhhhhh I see what happened, their solution, for the total cases, it only uses the cases $(a,b)$ where a is smaller than b, so there are only 10, and there is only 1 number larger than 2 that divides it 4, i thought there was a 6 for some reason, so the final probability is $\frac{5}{10}=\boxed{\frac{1}{2}}$

@weredumbmaster22, you have an incomplete question: It stops at "is 36, then how long"-with an abrupt stop. Next time you post a question, please check and make sure everything is complete.

Start with the smaller one:

you have 1, which has 4 larger numbers that can divide it

you have 2, which has 2 larger numbers that can divide it

you have 3, which has 0 larger numbers that can divide it

you have 4, which has 0 larger numbers that can divide it

you have 5, which has 0 larger numbers that can divide it

Lastly, there are $5^2=25$ total combinations, and only 6 of which work. Our end result is $\boxed{\frac{6}{25}}$