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What is the nth term of the sequence 6 0 -6 -12?
There are infinitely many polynomials of order 4 (or higher) that will give these as the first four numbers and any one of these could be "the" rule. Short of reading the mind of the person who posed the question, there is no way of determining which of the infinitely many solutions is the "correct" one.The simplest rule is a linear polynomial U(n) = 6*(2 - n) for n = 1 , 2, 3, ...
Why you can you do n-12 but not 12-n?
The subtraction is not commutative, it means that n - 12 is not the same as 12 - n, except when n is 12. For example, suppose that we have a statement such as n - 12 = 12 - n. Can we find a value for n which will satisfy our statement?n - 12 = 12 - n (add n and 12 to both sides)n + n - 12 + 12 = 12 + 12 - n + n2n = 24 (divide by 2 both sides)n = 12Check:n - 12 = 12 - n12 - 12 =? 12 - 120 = 0 TrueBut, for all other values of n, the statement is not true. Let's say that n = 20Does 20 satisfies the statement? Check:n - 12 = 12 - n20 - 12 =? 12 - 208 = -8 FalseHence, we cannot write n - 12 as 12 - n for n ≠12.
Find three consecutive odd integers such that the product of the first and the second exceeds the third by 8?
Let first COI=n=3 second COI=n+2=5 third COI=n+4=7 (n)(n+2)=n+4+8 check: n^2+2n=n+12 n^2+n-12=0 -n-12 -n-12 -4^2+(-4)-12=0 n^2+2n-n-12=0 16+(-4)-12=0 n^2+n-12=0 12-12=0 (n+4)(n-3)=0 0=0 n+4=0 n-3= 0 n^2+n-12=0 -4 -4 +3 +3 3^2+3-12=0 n=-4 n=3 9+3-12=0 reject 12-12=0 because is negative and it needs to be positive 0=0
-3(n plus 5)12?
If you mean: -3(n+5) = 12 then n works out as -9 because -3(-9+5) = 12
How do you factor n squared minus 16n plus 48?
(n - 4)(n - 12)
What is '12 N on a C F ' please?
12 Numbers on a Clock Face
How many different ways are there to choose a dozen donuts from the 21 varieties at a donut shop?
The answer depends on if you can choose the same kind of donuts more then once. Or in other words, is repetition permitted. If you can only choose the same kind of donuts only once, it is a 21 choose 12 problem: C(n,k) = n! / (k! (n - k)!) C(21, 12) = 21! / (12! (21 - 12)!) = 21! / (12! (9)!) = 293,930 If you can choose the same kind of donuts more then once, it is a combination with repetition problem. P(n+k-1,k) = (n+k-1)! / (k! (n-1)!) or put it into C(n,k) with n+k-1 as 21 + 12 - 1 = 32 and k as 12 so C(21+12-1,12) = C(32, 12) = 32! / (12! (32 - 12)!) = 32! / (12! (20)!) = 225,792,840
12 N on a C F?
12 Noon on a clock face. 12 Numbers on a clock face.
What elements are there in B-12?
Vitamin B-12 contain Co, C, H, O and N.
What is the scope of a general statement of an infinite surd?
I think that you're doing the same thing that I did for IB. Here's what I think you're asking for : 0, 2, 6, 12, 20, 30, 42... c₁ = 0 c₂ = 2 c₃ = 6 (c₂ + 4 = 2 + 4 = 6) c₄ = 12 (c₃ + 6 = 2 + 4 + 6 = 12) c₅ = 20 (c₄ + 8 = 2 + 4 + 6 + 8 = 20) c₆ = 2 + 4 + 6 + 8 + 10 etc... dn = (n/2) <2c₁ + (n-1) 2> dn = (n/2) <2 (2) + (n-1) 2> dn = (n/2) (4 + 2n - 2) dn = (n/2) (2 + 2n) dn = (2n/2) + (2n²/2) dn = n + n²
Does wheel have 2 syllables?
Neither it has 12 You c u n t
What is the nth term of the sequence 6 0 -6 -12?
There are infinitely many polynomials of order 4 (or higher) that will give these as the first four numbers and any one of these could be "the" rule. Short of reading the mind of the person who posed the question, there is no way of determining which of the infinitely many solutions is the "correct" one.The simplest rule is a linear polynomial U(n) = 6*(2 - n) for n = 1 , 2, 3, ...
Which has more protons in the nucleus C-12 C-14 or N-14?
N-14 has more protons. It contains one more proton than C.
What city is located at 42 N 12 E?
c
Create a program that will ask the user to input an integer and display all odd numbers?
HERE IS IT!!!!!#include#includevoid main(){int a[12],b[12],c[12],i,j=0,k=0;clrscr();printf("Enter an integer\n");for(i=0;i
12n to the fifth power divided by 36n equals?
0.3333
How many letters are there in ancient Egypt?
A-N-C-I-E-N-T-E-G-Y-P-T {12 Letters}
