cot x = (cos x) / (sin x) cos (x - 180) = cos x cos 180 + sin x sin 180 = - cos x sin (x - 180) = sin x cos 180 - cos x sin 180 = - sin x cot (x - 180) = (cos (x - 180)) / (sin (x - 180)) = (- cos x) / (- sin x) = (cos x) / (sin x) = cot x
60 of 180 = 60 x 180 = 10,80060% of 180 = 180 x 0.6 = 10860% off 180 = 180 x 0.4 = 7260 off 180 = 180 - 60 = 120
To get the answer to this question you have to multiply 180 by 24 to get the number of hours. So 180 x 24 = 4320. So the answer to your question is, there are 4320 hours in 180 days.
55 + x = 180 x = 180 - 55 x = 125
180 x 40 = 7200
2 x 2 x 3 x 3 x 5 = 180 2 x 5 x 7 = 70 23 x 32 x 52 x 7 = 12,600 = 180 x 70
84 = 2 x 2 x 3 x 7; 180 = 2 x 2 x 3 x 3 x 5; Common factors are 2 x 2 x 3 so LCM = 7 x 180 = 1260
The prime factorization of 180 is 2 x 2 x 3 x 3 x 5. The prime factorization of 70 is 2 x 5 x 7. The prime factorization of 180 x 70 is 2 x 2 x 2 x 3 x 3 x 5 x 5 x 7 or 23 x 32 x 52 x 7
The LCM of 24, 42, and 180 = 2520. 24 = 2^3 x 3 42 = 2 x 3 x 7 180 = 2^2 x 3^2 x 5 lcm = 2^3 x 3^2 x 5 x 7 = 2520
There are 180 days in a school year. Most schools go for 7 hours. If you multiply 180 x 7 x 11 you get 13860
60 x 80 = 4800 60 x 3 = 180 80 x 7 = 560 3 x 7 = 21 4800 + 180 + 560 = 21 = 5561 67 x 83 = 5561
15 times with a remainder of '7'. 15 x 12 = 180 180 + 7 = 187
7
180 = 22 x 32 x 5 630 = 2 x 32 x 5 x 7 So the common prime factors are 2, 32 & 5 The hcf of 180 and 630 = 2 x 32 x 5 = 90 (take lowest power of all primes) The LCM of 180 and 630 = 22 x 32 x 5 x 7 = 1260 (take highest power of all primes)
cot x = (cos x) / (sin x) cos (x - 180) = cos x cos 180 + sin x sin 180 = - cos x sin (x - 180) = sin x cos 180 - cos x sin 180 = - sin x cot (x - 180) = (cos (x - 180)) / (sin (x - 180)) = (- cos x) / (- sin x) = (cos x) / (sin x) = cot x
592 6 x 7 = 42 6 x 30 = 180 10 x 7 = 70 10 x 30 = 300 Total: 592
Here's the factor tree: 180 =90x2 =45x2x2 =15x3x2x2 =5x3x3x2x2 or 2 x 2 x 3 x 3 x 5 (In index form this is 22 x 32 x 5)