jimkey17

We know that this is a right triangle since it has pythagorean triples 

see here:

https://www.geogebra.org/calculator/hty44jpe

I plotted here

we can use algebra to form the equation 25x+5(x+16)=590 we distribute to get 25x+(5)(x)+(5)(16)=590 then it becomes 25x+5x+80=590 and we combine like terms to get (25x+5x)+(80)=590 then 30x+80=590 we then subtract 80 from both sides to get 30x+80−80=590−80 then 30x=510 finally we divide both sides by 30 to get 30x/30=510/30 then we have x=17 so we get that she has \(17\text{ Quarters}\)

and 33 nickels :)

To check we do 5*33+17*25 which simplifies into 165+425=590 so that means that our answer 

AREA:   10*8/2 for the area of the triangle and π4^2 so we add them up in a equation to get that the area 80/2+π4^2=40+16π so the area of the shaded region is 40+16π or \(\boxed{40+16π}\)

Circumfrence=8π/2 for the rounded part and 2*(sqrt(10^2+(8/2)^2)) again we make them into an equation 4π+2*(sqrt(10^2+4^2))=4π+2*(sqrt(116)) then we simplify it to get 4π+4*(sqrt(29)) or \(4π+4\sqrt{29}\)

We know that the perimeter is equal to 2x-1+2x-1+3x+3+3x+3=144 which can be simplfied to 4x-2+6x+6=114 and then 10x+4=114 we then subtract 4 to get 110 and divide by 10 to get x=11 so PQ=11*3+3=33+3=36 and we have that PQ=36 or \(\boxed{PQ=36}\)

let's say that y=-0.5 since that works with -1 < y < 0 so we compute that  A=0.25, B=1.5, C=0.5, D=-1 so we eventually see that B is the greatest so we get B or \(\color{green}{\boxed{B}}\)

an angle in a regular pentagon is 108^o and the triangle is isosceles so the measure of DEB is 108-((180-108)/2)=108-(72/2)=108-36=72^o sp we have that \(\angle DEB\) is 72^o or \(\boxed{72^\circ}\).

Thanks Guest I forgot the final step

\(z-\sqrt{-2i},z=-\sqrt{2i},z=\sqrt{-2i},\text{ and }z=\sqrt{2i}.\) since we quad root to get \(\pm\sqrt{\pm2i}\)

We have C(25,3)=2300 possible ways and there are C(3,1)*12=3*12=36 ways so the probability that at least two of the three had been sitting next to each other is \(\dfrac{36}{2300}=\dfrac9{575}\)

that means that we can seperate the hexagon into 18 equal traignles with each area 1/3 and muliply it 12 since there are 12 of the triangles that fit exactly in the star so the area of the star is 12*1/3=4