\(\angle N=\angle O=90\)
\(\angle J+\angle H=360-\angle O-\angle N=360-90-90=180\)
I'm not sure about this
(2x - 30) + x = 180
3x = 210
x = 70
arc CB = 2x - 30 = 140 - 30 = 110
\(4125=3\times 5^3\times 11\) first 3 multiple of 5 : 5,10,15
m = 15
\(2816000=2^{11}\times 5^3\times 11\) multiple of 2 : \(2, 2^2, 2\times 3, 2^3, 2\times 5, 2^2\times 3, 2\times 7\)
n=15
n - m = 15 - 15 = 0
9z + 7 + 74 = 180
9z = 99
z = 11
\(\angle A+\angle C=180\)
\(\angle A+101=180\)
\(\angle A=79\)
\(\angle B+\angle D=180\)
\(\angle B+68=180\)
\(\angle B=112\)
(4n - 2) + (3n + 8) = 180 - 90
7n + 6 = 90
7n = 84
n = 12
(2y + 1) + (5y - 3) = 180
7y - 2 = 180
7y = 182
y = 26
volume = \((6\times 3\times 1)+(4\times 3\times 8)=18+96=114ft^3\)
Volume = \(\frac{1}{3}\times (4\times 10)\times 6=2\times 40=80 cm^3\)
\(300=2^2\times 3\times 5^2\)
Because A the product of the divisors of 300 who only have this 3 distinct prime divisors, so there are 3 distinct prime divisors from A.
