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Base units need not have ANY volume. A second is the base unit for measuring time and it has no volume!
Volume = Base Area * Height So Height = Volume/Base Area
Divide the base area into its volume
Volume of a rectangular prism is equal to the base area x height. (V=BxH or Volume = Base Area * Height) To get the base area, simply divide the volume by the height. (B=V/H or Base Area = Volume / Height)
Base = sqrt(Volume/Height)
That can't be answered. The volume of the base of what kind of base?
The number of digits is immaterial, the principal is exactly the same. The key to reversing any number is in the following user-defined function: uint reverse(uint num) { uint rev=0; while(num) { rev*=10; rev+=num%10; num/=10; } return(rev); } The following code demonstrates how to reverse a 6 digit number entered from the console. #include<iostream> #include<string> #include<sstream> typedef unsigned int uint; uint reverse(uint num) { uint rev=0; while(num) { rev*=10; rev+=num%10; num/=10; } return(rev); } int getrange(uint digits,uint& min,uint&max) { if(!digits) return(-1); min=1; while(--digits) min*=10; max=min*10-1; return(0); } uint inputnumber(const std::string prompt,const uint min,const uint max) { using namespace std; uint result=0; while(1) { cout<<prompt; string s; getline(cin,s); if(s.size()) result=stoul(s,0,10); if(result<min result>max) cout<<"The number must be in the range " <<min<<".."<<max<<"\n" <<"Please enter another number.\n"<<endl; else break; } return(result); } uint inputnumber(const uint digitcount) { using namespace std; uint min=0, max=0; if(getrange(digitcount,min,max)==-1) return(0); std::stringstream prompt; prompt<<"Enter a "<<digitcount<<" digit number: "; return(inputnumber(prompt.str(),min,max)); } int main() { using namespace std; uint num=inputnumber(6); // 6 digits. if(num) { uint rev=reverse(num); cout<<"Reverse: "<<rev<<"\n"<<endl; } return(0); } Output: Enter a 6 digit number: 123456 Reverse: 654321
You mean an iterative function, not a recursive one. A recursive function is one that calls itself with altered arguments, typically used to implement divide-and-conquer algorithms. Since the arguments remain the same and you're not dividing the array into smaller subsets, recursion is not necessary. An iterative loop is all you need: typedef unsigned int UINT; const UINT CountOddNumbers( const UINT* const A, const UINT size) { UINT result=0; for(UINT i=0; i<size; ++i) result += A[i]&1; return(result); } Note that this function is not safe, since size is not guaranteed to reflect the actual size of A. In C++, it's better to use a vector object, since vectors encapsulate the actual size of the embedded array: const UINT CountOddNumbers( const std::vector<UINT> A) { UINT result=0; for(UINT i=0; i<A.size(); ++i) result += A[i]&1; return(result); }
Base units need not have ANY volume. A second is the base unit for measuring time and it has no volume!
Volume = Base Area * Height So Height = Volume/Base Area
Well,i guess the same volume as the base.
volume=base multiplied by heigth
Divide the base area into its volume
Volume of a rectangular prism is equal to the base area x height. (V=BxH or Volume = Base Area * Height) To get the base area, simply divide the volume by the height. (B=V/H or Base Area = Volume / Height)
Volume of a rectangular prism = base x height. If volume and height are known, solve for base area by dividing volume by height.
Volume = base area x height so Base area = Volume / height. For example, if the volume is 10cm3 and the height is 2cm, 10/2=5, the base area must be 5cm2.
Since: Volume = height x base Height = Volume / Base