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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 $\binom{17}{7}=19448$

My attempt

All pronumerals are integers

LaTex not displaying so here is a pic that I took of it.

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\(0

which means there are 18 possible vaues of g and hence 18 possible values of n

LaTex:

0 4n=2\mod6\\
2n=1\mod3\\
2n=3k+1\\
2n-3k=1\\
\text{One solution is n=2, k=1 }\\
2(2\qquad)-3(1\qquad)=1\\
2(2+3g)-3(1+2g)=1\\
so\\
n=2+3g\\
0<2+3g<60\\
-2<3g<58\\
-0.6 0\le g\le17

180 - 120 = 60

x + 65 + 60 = 180

x + 125 = 180

x = 55

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.

0 < n < 60

\(4n\equiv 2\pmod 6\)

\(n\equiv \frac{2}{4}\pmod6\)

\(n\equiv \frac{2+6}{4}\equiv2\pmod6\)

n = 6k + 2

k = 0, 1, 2, ..., 9

total : 10

\(\frac{2}{3}\times m=w\)

\(\frac{5}{8}\times \frac{1}{3}\times m=\frac{5}{24}\times m=b\)

\(m(1-\frac{2}{3}-\frac{5}{24})=6\) => \(\frac{1}{8}\times m=6\) => \(m=48\)

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)

oh wait thank you so much. i totally forgot about the parentheses and didn't distribute the square to the 2 also.

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 $\frac{5}{10}=\boxed{\frac{1}{2}}$

600 ml x 0.7 = 420 ml of alcohol is in the solution. 

no... i sadly got this wrong. the solution it gave me:

There are 10 pairs of integers that we can potentially select. The easiest way to do this is to simply write them all out: (1,2), (1,3), (1,4), (1,5), (2,3), (2,4), (2,5), (3,4), (3,5), and (4,5). The 4 pairs with 1 as an element all obviously work, as does (2,4), but none of the others do. That means that 5 out of 10 pairs work which gives us a probability of 1/2. thank you for helping, though! :)

Like I understand that you have to find the intervals, but I'm not sure how to solve the inequalities for them...
Three cases: (at least I think)

\(y<\frac{-1}{4} \)     and.    \(-\frac{1}{4} ≤ y < \frac{3}{2}\)     and     \(y ≥ \frac{3}{2}\)

p.s. why are only some of them like "≤ and ≥" but others aren't? 

it's just very confusing to me...
 

@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 $\boxed{\frac{6}{25}}$

\(-99 - \sqrt{7801}= {-b - \sqrt{b^2-4ac} \over 2a}\)

\(-99 - \sqrt{7801}= {-99 - \sqrt{99^2-4(\frac{1}{2})c} \over 2 \cdot \frac{1}{2}}\)

\( \sqrt{7801}= \sqrt{99^2-4(\frac{1}{2})c}\)

\(7801 = 9801 - 2c\)

\(2c = 2000\)

\(c = 1000\)

hope this helps :)

By the Pythagorean Theorem, the side length is 35.

CC2 = 30*sqrt(5).

3/2   *  2/3   = 1   daily value of Vit C

1+i      1 -i      -1 +i      -1 -i

Thanks. Sorry I didn't mean to name this "probability" lol

Hint: The $y$ coordinate turns into $4-(y-4)=8-y$. Now put $y\mapsto 8-y$ in your equation and expand :)

Related rates and the pythagorean theorem

x = car 1 = 2 hrs * 10 m/hr = 20 miles     and      dx/dt = 10  m/hr

y = car 2 = 2hrs * 20 m/hr = 40 miles      and      py/dt = 20 m/hr

Pythagorean theorem for the distance between the two cars :

x^2 + y^2 = s^2                        s^2 = 2000      now differentiate with respect to time (using chain rule)

2x dx/dt     +    2y dy/dy   =   2s ds/dt           We have all of these variables except   ds/dt (which we are looking for) ...so plug in

2(20)  * 10          + 2 (40)  *20  =   2 (sqrt(2000)  * ds/dt

ds/dt = 2000/(2 sqrt(2000))  = 1000/sqrt(2000) = 22.36 m/hr

Given that x,y, and z are all integers, we have $\frac{2}{x}=1$ or $2$. So we have $x=1$ or $x=2$. Symmetrically, the same is true for $y$ and $z$. So we have $1+1+1$, $2+1+1$, $2+2+1$, and $2+2+2$ as our only distinct options. Counting, we see there are $\boxed{4}$ of them.

It's not a glitch, it is because someone previously spammed on here, and then I flagged the post and the admin came on and took down the spam posts. There were 2 spam posts and 2 of my replies. They are still there, just only visible to the mods and the admin.