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What is the name of the sequence -3 -2 -1 0 1 2?
The set of integers.
Which set best represents the integers?
.{..., -3, -2, -1, 0, 1, 2, 3, ...}
What is the numerical value of a 10 bit word?
presuming you mean ten bits, every bit is a power of 2, beginning with power 0, right to left. So, 11 = 2^1+2^0=3 and 101=2^2+0+2^0=5 The value of a ten bit word depends on the bits which are set (=1). This allows representation of values from 0 through 1023, effectively 1K as rounded in "computerese". All bits set=1 yields 1023, no bits set (all clear) yields 0.
Which set of ordered pairs corresponds to the graphWhich set of ordered pairs corresponds to the graph 5 1 2 3 0 1 4 2 5 1 2 4 1 5 1 0 1 5 3 2?
It would help if we could see the said graph but this browser is simply not up to it.
How many different equations can be made with the numbers 0123?
A huge number. 0 + 1 + 2 = 3 0 + 2 + 1 = 3 1 + 0 + 2 = 3 1 + 2 + 0 = 3 2 + 0 + 1 = 3 2 + 1 + 0 = 3 -0 + 1 + 2 = 3 -0 + 2 + 1 = 3 1 - 0 + 2 = 31 + 2 - 0 = 32 - 0 + 1 = 32 + 1 - 0 = 3 0 - 1 + 3 = 2 0 + 3 - 1 = 2 -1 + 0 + 3 = 2 -1 + 3 + 0 = 2 3 + 0 - 1 = 2 3 - 1 + 0 = 2 -0 - 1 + 3 = 2-0 + 3 - 1 = 2-1 - 0 + 3 = 2-1 + 3 - 0 = 23 - 0 - 1 = 23 - 1 - 0 = 2 0 - 2 + 3 = 1 0 + 3 - 2 = 1 -2 + 0 + 3 = 1 -2 + 3 + 0 = 1 3 + 0 - 2 = 1 3 - 2 + 0 = 1 -0 - 2 + 3 = 1-0 + 3 - 2 = 1-2 - 0 + 3 = 1-2 + 3 - 0 = 13 - 0 - 2 = 13 - 2 - 0 = 1 1 + 2 - 3 = 0 1 - 3 + 2 = 0 2 + 1 - 3 = 0 2 - 3 + 1 = 0 -3 + 1 + 2 = 0 -3 + 2 + 1 = 0 For each of these equations there is a counterpart in which all signs have been switched. For example 0 + 1 + 2 = 3 gives -0 - 1 - 2 = -3and so on. Now, all of the above equations has three numbers on the left and one on the right. Each can be converted to others with two numbers on each side. For example:the equation 0 + 1 + 2 = 3 gives rise to0 + 1 = 3 - 20 + 1 = -2 + 30 + 2 = 3 - 10 + 2 = -1 + 31 + 2 = 3 - 01 + 2 = -0 + 3-0 + 1 = 3 - 2-0 + 1 = -2 + 3-0 + 2 = 3 - 1-0 + 2 = -1 + 31 + 2 = 3 + 01 + 2 = +0 + 3 As you can see, the number of equations is huge!
How do you write a C program to print all combinations of a 3-digit number?
Mathematically-speaking there is only one combination of a 3-digit number. When dealing with a combination of digits, the order of those digits does not matter, thus 123 and 321 are the exact same combination. If the order of the digits is important then it is a permutation, not a combination. That is, 123 and 321 are completely different permutations of the same combination of digits. The confusion is understandable given that we commonly refer to a combination lock instead of a permutation lock. Mathematically speaking, a combination lock that unlocks with the code 123 would also unlock with the codes 132, 213, 231, 312 and 321, because all six permutations of the digits 1, 2 and 3 are in fact the same combination. Some combination locks really do work this way, however the vast majority are actually permutation locks; we call them combination locks simply because we don't normally use the term "combination" in the much stricter mathematical sense. To restate the question: How do you write a C program to print all permutations of a 3-digit number? A 3-digit number has 6 permutations, thus we can print all six by treating the number as an array, and printing all 6 permutations of the array: void print_permutations (int num) { char set[3]; int index; if (num<100 num > 999) return; // not a 3-digit number index = 0; while (num>0) { set[index] = num % 10; // least-significant digit num /= 10; // shift all digits one position to the right } // sort the array in ascending order if (set[0]>set[1]) set[0]^=set[1]^=set[0]^=set[1]; if (set[1]>set[2]) set[1]^=set[2]^=set[1]^=set[2]; if (set[0]>set[1]) set[0]^=set[1]^=set[0]^=set[1]; // print the permutations printf ("%d%d%d\n", set[0], set[1], set[2]); printf ("%d%d%d\n", set[0], set[2], set[1]); printf ("%d%d%d\n", set[1], set[0], set[2]); printf ("%d%d%d\n", set[1], set[2], set[0]); printf ("%d%d%d\n", set[2], set[0], set[1]); printf ("%d%d%d\n", set[2], set[1], set[0]); } The problem with this is when the 3-digit number contains duplicate digits. This would treat 100 as if it had 6 permutations when it really only has 3 {100, 010 and 001}, while 111 only has one permutation {111}. These must be treated as being special cases: void print_permutations (int num) { char set[3]; int index; if (num<100 num > 999) return; // not a 3-digit number index = 0; while (num>0) { set[index] = num % 10; // least-significant digit num /= 10; // shift all digits one position to the right } // sort the array in ascending order if (set[0]>set[1]) set[0]^=set[1]^=set[0]^=set[1]; // move larger of 1st and 2nd digit to middle if (set[1]>set[2]) set[1]^=set[2]^=set[1]^=set[2]; // move larger of 2nd and 3rd digit to end if (set[0]>set[1]) set[0]^=set[1]^=set[0]^=set[1]; // move larger of 1st and 2nd digit to middle // print the permutations (handle special cases where digits are duplicated) if (set[0]==set[1] && set[0]==set[2]]) { // same three digits (one permutation) printf ("%d%d%d\n", set[0], set[1], set[2]); } else if (set[0]==set[1]) { // first two digits are the same (three permutations) printf ("%d%d%d\n", set[0], set[1], set[2]); // same as 1, 0, 2 printf ("%d%d%d\n", set[0], set[2], set[1]); // same as 1, 2, 0 printf ("%d%d%d\n", set[2], set[0], set[1]); // same as 2, 1, 0 } else if (set[1]==set[2]) { // last two digits are the same (three permutations) printf ("%d%d%d\n", set[0], set[1], set[2]); // same as 0, 2, 1 printf ("%d%d%d\n", set[1], set[0], set[2]); // same as 2, 0, 1 printf ("%d%d%d\n", set[2], set[1], set[0]); // same as 1, 2, 0 } else { // all three digits are unique (6 permutations) printf ("%d%d%d\n", set[0], set[1], set[2]); printf ("%d%d%d\n", set[0], set[2], set[1]); printf ("%d%d%d\n", set[1], set[0], set[2]); printf ("%d%d%d\n", set[1], set[2], set[0]); printf ("%d%d%d\n", set[2], set[0], set[1]); printf ("%d%d%d\n", set[2], set[1], set[0]); } }
The set ...-4 -3 -2 -1 0 1 2 3 4... is called the set of?
The set of integers.
What is the solution set for 2 cos 0 - 1 equals 0?
The fisrt thing to note is that there is no variable in the question and so there cannot be a solution set. The only possibilities are the statement is true, false or indeterminate. Now, 2 cos 0 - 1 = 0 is equivalent to 2*1 - 1 = 0 [since cos(0) = 1] or 2 - 1 = 0 which is false.
How can the median in a set of numbers be 0?
the median is the middle in a set of numbers so if you had a set of 5 numbers like -2, -1, 0, 1, and 2 the middle number would be zero.
What is the name of the sequence -3 -2 -1 0 1 2?
The set of integers.
What is the quantum number set for Oxygen?
The quantum number set for oxygen is n=2, l=0, ml=0, ms= +1/2.
How is 1 the generator of the set of integers?
1+1=2, 2+1=3, 3+1=4, ... 1-1=0, 0-1=-1, -1-1=-2, -2-1=-3, ...
What is the solution set for 2p2-5p plus 2 equals 0?
2p2-5p+2 = 0 (2p-1)(p-2) = 0 p = 1/2 and p = 2
Is 0.72 an integer?
No, the set of integers is {..., -3, -2, -1, 0, 1, 2, 3, ...}.
Which set best represents the integers?
.{..., -3, -2, -1, 0, 1, 2, 3, ...}
The set of counting numbers the opposites of the counting numbers and zero describes the set of?
The set of integers I. I = {..., -3, -2, -1, 0, 1, 2, 3, ...}
What is the solution set 2x2-3x-1 equals 0?
(-2,-1)
