Perimeter = 2L + 2W 350 = 2(93) + 2W 350 = 186 + 2W 350 - 186 = 186 - 186 + 2W 164 = 2W 164/2 = 2W/2 82 = W So the width is 82 cm.
60.0625 square feet.
The diameter is 93.
The diameter is 29.6
In any triangle whose sides have measures a,b, and c, the triangular inequality tells us: a + b > c a + c > b b + c > a Since the triangles here have two sides with the same measure, we will suppose that a = b and the above becomes a + a > c a + c > a a + c > a or 2a > c c > 0 c > 0 So we have 1) 2a > c > 0 Since the perimeter is 93, we have a + a + c = 93 2) 2a + c = 93 Solve 2) for c, c = 93 - 2a Substitute in 1) 1) 2a > 93 - 2a 4a > 93 a > 23.25 and since a is an integer a > 23 Solve 2) for 2a 2a = 93 - c Substitute 93 - c for 2a in 1) 93 - c > c 93 > 2c 46.5 > c and since c is an integer, 47 > c or c < 47 So we have a > 23 c < 47 Let p be the positive integer amount that a is greater than 23 Let q be the positive integer amount that a is less than 47 So a = 23 + p and c = 47 - q Substituting those into 2) 2a + c = 93 2(23 + p) + (47 - q) = 93 46 + 2p + 47 - q = 93 93 + 2p - q = 93 2p - q = 0 2p = q So now we have a = 23+p and c = 47-2p Since c > 0 47 - 2p > 0 -2p > -47 p < 23.5 and since p is an integer p < 24 And p > 0 so 0 < p < 24 So there are 23 values p can take on, any of the integers which are greater than 0 and less than 24. That's 1 through 23, so there are 23 possible isoceles triangles with integer sides that have perimeter 93. Here they all are: p a = 23+p b = a c = 47-2p ---------------------------------- 1. 24 24 45 2. 25 25 43 3. 26 26 41 4. 27 27 39 5. 28 28 37 6. 29 29 35 7. 30 30 33 8. 31 31 31 9. 32 32 29 10. 33 33 27 11. 34 34 25 12. 35 35 23 13. 36 36 21 14. 37 37 19 15. 38 38 17 16. 39 39 15 17. 40 40 13 18. 41 41 11 19. 42 42 9 20. 43 43 7 21. 44 44 5 22. 45 45 3 23. 46 46 1 Edwin
If you define a rectangular number as a number which is the product of two consecutive integers, none of them qualify.
No,but 120 is and 301 isn't
If you define a rectangular number as a number which is the product of two consecutive integers, none of them qualify.
If you define a rectangular number as a number which is the product of two consecutive integers, none of them qualify.
A rectangular number is any number greater than or equal to 2, that is the product of an positive integer n multiplied by the integer that comes before it. So 2x1 for example or 5x4 or 10x111. If you look at the number 93120301=n(n+1), this number is rectangular if n^2+n-93120301=0 has a solution that is a positive integer. There is no integer solution so that number is not rectangular. How about 93? n^2+n-93=0 also has no solution in the natural numbers.
The GCF of 93 and 120 is 3.
Factors of 120... 1,2,3,4,6,10,12,20,30,40,60,120 - Factors of 93... 1,3,31,93... therefore the GCF is 3
The LCM of 52, 93, and 120 is 3720
0.2583
108
93
93 is composite, 3x31=93