shmewy

Trisect each side of an equilateral triangle of side length 2 and construct an equilateral triangle on the center third of each side. Omit the side of each new triangle that was once a part of the original triangle. Repeat the trisection, construction, and omission steps on each segment formed above, creating the figure below. When this process is repeated indefinitely, a new figure is approached. The square of the area of this new figure can be written as p/q
where p and q are relatively prime positive integers. Find p + q. ps. the figure is a koch flake. pretty sure the answer is 223, but i some conformation. thank you :)

this is the edited version. thank u :)
 

$\begin{align*} \dfrac{s}{2} + 5t &= 3,\\ 3t - 6s &= 9. \end{align*}$

so ill first multiply the top equation by twelve to get 6s+60t=36

ill rearange the second equation to get -6s+3t=9

ill "add" the two equations to cancel out the s to get 63t=45

t=5/7

ill substitute t back into the first equation to get s/2+25/7=3

multiplying by two gets s+50/7=6

s=6-50/7

s=42/7-50/7

s=-8/7

so the ordered pair $(s,t)$would be $\boxed{(-\frac{8}{7},\frac{5}{7})}$

$127 \equiv 18 \pmod{n}$

so 109 = 0 mod n

109 is prime so the only two solutions are $\boxed{1\space and \space 109}$

The only one that is 2 mod 7 is 2.

so if a 83 degree sector of a circle has an area of 60 pi, then we can find some ratios. 83/360 = 60pi/ (area of cricle) so the area of the cirlc ewould be 21600/83 pi so the radius of the circle would be 60sqrt6 / sqrt83 or $\frac{60\sqrt{498}}{83}$. 

The equilateral triangles are in total two and a half units across which means the area of the square is 2.5^2 or $\boxed{\frac{25}{4}}$. and why so much geometry :P

to find the radius of the circle imagine a point M as the midpoint of QP. MC would then be the radius of the semicircle. MP is 8 and CP is 16 so we can find the hypotenuse by doing 64+256 then square rooting sqrt320 is $\boxed{8\sqrt[]{5}}$

the area of the trapezoid would be (34 +16)/2 * 16 or 25*16 or 100*4 or $\boxed{400}$ that wasnt so bad... :P

are you sure that the question is even valid because this is what it looks like 

we can solve this problem by looking at right triangle QVU. ST:PQ is 5:7. so QU is 7/12 of SQ. QV would be 7/12 of QR. the problem didn't give us QR or QS so we can make up values for them, because they would eventually cancel out. so ill set QR to 48, and QS would therefore be 52. after doing the calculations we get that QU is 91/3 and QV is 28. Applying the pythagorean thereom we get the UV is $\boxed{\frac{35}{3}}$.

$x^2+y^2=2x-6y+6+14x-14y+28$

This is just a circle. i moved the xs and ys to the left and that just moves the circle. the size depends on the constant which is 34. so the area of the circle would be 34pi

Try to use polynomial division by dividing values of (x +or- $\frac{divisors\space of \space 12}{divisors\space of \space 10}$)

im wayyyyy too lazy to do it. but eventually by trying all values you will get a quadratic. then it becomes mucchhhh easier :)

$x^2 + nx + 15 < -89 - 3x^2$

so i first brought everything to the left side to get $4x^2 + nx + 104 < 0$

then i divided by 4 to get$x^2 + \frac{n}{4}x + 26 < 0$

for x to have no real solutions the discriminant must be negative

so $\frac{n^2}{16}-4(26)$ must be negative. the less than sign is useless since smaller right hand values make c greater

the greatest value of n that works is 40

but we also can have negatives, least of which is -40

so -40 to 40 is $\boxed{81}$ numbers

$(x^2 + 1/x)*(x + x^4)^3$

i then reformatted stuff to get$(\frac{x^3+1}{x})(x^3+3x^6+3x^9+x^{12})$

i further expanded to get$\frac{x^3+4x^9+6x^9+4x^{12}+x^{15}}{x}$

then just substract one from all the powers in the numerator to get the final answer of $\boxed{x^2+4x^5+6x^8+4x^{11}+x^{14}}$. :)

XD just expanding is definentaly(100% misspelled) a way to do it