2(x+80+70)=360=2x+300
x=$\boxed{30}$
By power of point, we have 10$\cdot$27=9(9+x)=270=81+9x
9x=189
x=$\boxed{21}$
1. y=5$\sqrt2$
x=10
2. y=4$\sqrt6$
x=8$\sqrt6$
3.x=5$\sqrt3$
4. 12$\sqrt3$
$\boxed{B}$ because 70+110=180, which is true.
2x2 1/4=9/2
(9/2)/(1/2)=9/2x2/1=$\boxed{9}$
Lets call BD x. using pythagorean theorum, we have:
16+x^2+225+x^2=361
241+2x^2=361
x^2=60
x=$\boxed{2\sqrt{15}}$
5^8+5^7+5^6+5^5+5^4+5^3+5^2+5=$\boxed{488280}$
(180-136)/2=22
10x-8=22
10x=30
x=$\boxed3$
105=3x7x5
3+x=4(mod 27)
5+x=9(mod125)
7+x=25(mod343)
x=1(mod27)
x=4(mod125)
x=18(mod343)
27k+1=x
3(9k)+1=x
x=1(mod3)
for the same reasons:
x=4(mod5)
x=4(mod7)
x=4(mod 35)
possible sollutions lower than 105:
4, 39, 74.
the only one that has a remainder 1 when divided by 3 is $\boxed{4}$
Cos (M) = $\frac{24^{2}+51^{2}-45^{2}}{2\cdot 24\cdot 51}=\frac{1152}{2448}$
M = cos$^{-1}$($\frac{1152}{2448}$) ≈ 61.9 ≈ $\boxed{62}$
