5
Let n be the number, then: 1/n - 1/3n = 1/6 Solve the subtraction on the left: 1/n - 1/3n = 3/3n - 1/3n = 2/3n But we know this is 1/6, thus: 2/3n = 1/6 ⇒ 3n = 12 ⇒ n = 4 Thus the number is 4.
3n+1=7 3*2+1=7
The Series has the formula 3n + 1/2(n - 1)(n - 2) = 3n + 1/2(n2 - 3n + 2) which simplifies to, 1/2(n2 +3n + 2)
3n + 4 = 7(- 4 from both sides)3n = 3(divide both sides by 3)n = 1
(3n)(3n-1) = 3n * 3n - 3n * 1 Now, perform the multiplication: (3n * 3n) = 9n^2 (3n * 1) = 3n So, (3n)(3n-1) simplifies to: 9n^2 - 3n
3n^4 divided by 3n^3 = n
3n + 2 + (3n + 3 - 3n + 1) = 3n + 2 + (3n + 3 - 3n + 1) = 3n + 2 + (4) = 3n + 6
3n+2 + (3n+3 - 3n+1) = 3n+1+1 + (3n+1+2 - 3n+1) = 3*3n+1 + (9*3n+1 - 3n+1)= (3+9-1)*3n+1 = 11*3n+1.
The remainder is 0.If A has a remainder of 1 when divided by 3, then A = 3m + 1 for some integer mIf B has a remainder of 2 when divided by 3, then B = 3n + 1 for some integer n→ A + B = (3m + 1) + (3n + 2)= 3m + 3n + 1 + 2= 3m + 3n + 3= 3(m + n + 1)= 3k where k = m + n + 1 and is an integer→ A + B = 3k + 0→ remainder when A + B divided by 3 is 0-------------------------------------------------------------------------From this, you may be able to see that:if A when divided by C has remainder Ra; andif B when divided by C has remainder Rb; then(A + B) divided by C will have remainder equal to the remainder of (Ra + Rb) divided by C
5
Let n be the number, then: 1/n - 1/3n = 1/6 Solve the subtraction on the left: 1/n - 1/3n = 3/3n - 1/3n = 2/3n But we know this is 1/6, thus: 2/3n = 1/6 ⇒ 3n = 12 ⇒ n = 4 Thus the number is 4.
No. 3n isn't a factor of 3n + 7. The GCF of 3n + 7 and 9n is 1.
Thrice is three, "a number" can be shown as n. This is 3n. 3n/3=n
3n-2w and how do you work it out = 1
15-3n=n-1
In 3n-1 equals 11, n is 4.3n - 1 = 113n - 1 + 1 = 11 + 13n = 123n/3 = 12/3n = 4