I think that means sec (secant) or cos-1 or arccos
If x is one of the numbers and y the other, then their sum is x+y = 53. So y = 53-x where x>0 and y>0 implying that x<53. Then, their product, P = x*y = x*(53-x) = 53*x - x2 for 0<x<53 [P is not defined for other values of x]
Scary I had the same math problem with my teacher Mr. Dates, the answer is -53/50. This was on Lesson 2-10 I believe try using my.hrw.com next time.
53 x 53 = 2809
53 itself. It is a prime number.
Expressed as a percentage, 53/53 x 100 = 100 percent.
If x is one of the numbers and y the other, then their sum is x+y = 53. So y = 53-x where x>0 and y>0 implying that x<53. Then, their product, P = x*y = x*(53-x) = 53*x - x2 for 0<x<53 [P is not defined for other values of x]
Scary I had the same math problem with my teacher Mr. Dates, the answer is -53/50. This was on Lesson 2-10 I believe try using my.hrw.com next time.
Assuming the the numbers are generated by a cubic function, they are 107, 85 and 53
10% off of 53= 10% discount applied to 53= 53 - (10% * 53)= 53 - (0.10 * 53)= 53 - 5.3= 47.7
53% is a little over half. 53% of 100 is 53. 53% of 10 is 5.3. 53% of 200 is 106.
53 x 53 = 2809
56.18 100% of 53 = 53 6% of 53 = 0.06 x 53 = 3.18
1 and 53 are the factors of 53. 53 is a prime number, meaning only one and itself can be evenly divided into it.Factors of 53 are 53 & 1.
It is: -53/10 as an improper fraction
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33 percent of 53 = 17.4933% of 53= 33% * 53= 0.33 * 53= 17.49
30% off of 53 = 37.1= 30% discount applied to 53= 53 - (30% * 53)= 53 - (0.30 * 53)= 53 - 15.9= 37.1