jimkey17

n=12 since $365$<$\dbinom n4$ and $\dbinom {11}4$=330 and $\dbinom {12}4$=495 so n=12

T≈−24.945077 since we first subtract 0.000115T^3+0.000588t^2+1.732516t+1344.637096 from both sides.Then we simplify it into 1300−(0.000115t3+0.000588T2+1.732516T+1344.637096)=0.000115T3+0.000588t2+1.732516t+1344.637096−(0.000115t3+0.000588T2+1.732516T+1344.637096)−0.000115T3−0.000588T2−1.732516t−44.637096=0

Slide to see full explanation

for example the are 1 and x for the first one we get $\cfrac{\sqrt3}{4}$for the first area and for the height of the other triangle we have $\dfrac{2}{\sqrt3}$ for the base and 1 for the height so the ratio is $\dfrac{\sqrt3}{4}$ $:$ $\dfrac{2}{\sqrt3}$

so we get that z=$-\sqrt2$ since when we square it we get 2 and when we cube it we get $-2\sqrt2$ so we get... 

but there are no values of even number roots of negative numbers so there is no value to this solution

He already solved it

Well I got sqrt(5+2sqrt(6)) and sqrt(6) is about 2.44948974 and 2.44948974*2=4.89897948 also 4.89897948+5=9.89897948 then we solve sqrt(9.89897948) is about 3.14626437 so I agree that it is correct 

Hi could you please put this as off topic

Thank you

we simplify by subtracting 1 from both sides to get x=y-9 so we get that y is 9 more than x and also is 2 times x so y=-9 and x=-18

2)
so it is 0.6*0.82*2*[probability of the stickers pulled out in that order]