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Here's my attempt at a solution: int n=sizeof(arr)/sizeof(int); int size=2; int tempans[2]; . int answer[2];// the range is stored in another array. for(int i =0;i<n;i++) if(arr[0]<arr[1]) answer[0]=arr[0]; answer[1]=arr[1]; if(arr[1]<arr[0]) answer[0]=arr[1]; answer[1]=arr[0]; if(arr[i] < answer[1]) size += 1; else if(arr[i]>answer[1]) {
Arcsine's range comes from restricting sine's domain to $[-\pi/2,\ \pi/2]$ which covers the whole range of $[-1,1]$ and also includes the origin. Arcsecant's range comes from the arccosine's range. To make cosine invertible, the domain is restricted to $[0,\pi]$.
Use the graph to find the range. Interval Notation: [0, π 2)∪(π 2,π] [0, π 2) ∪ (π 2, π] Set -Builder Notation: {y∣∣0 ≤ y ≤ π,y ≠ π 2} {y | 0 ≤ y ≤ π, y ≠ π 2} Determine the domain and range. Domain: (−∞,−1]∪ [1,∞),{x|x ≤ −1,x ≥ 1} (- ∞, - 1] ∪ [1, ∞), {x | x ≤ - 1, x ≥ 1}
1 lut 2015 · Two obvious ways would be to provide the array and indices range (as lared mentioned) or specify the range using two pointers int ProfitB = Sum(array + A, array + A+B-1); /* sum array[A] ... array[A+B-1] */
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The range is the set of all valid y y values. Use the graph to find the range. Interval Notation: [0, π 2)∪(π 2,π] [0, π 2) ∪ (π 2, π] Set -Builder Notation: {y∣∣0 ≤ y ≤ π,y ≠ π 2} {y | 0 ≤ y ≤ π, y ≠ π 2}
ArcSecDegrees [z] gives the arc secant in degrees of the complex number z.
