It is simply: 4n as an algebraic term
(n+4)2
m^4 n^5 - m^20 n^21
9(4+n)Nine times the sum of 4 and a number is an expression and can not be evaluated.
It is 3*abs(n - 6) + 4*n
4(N-3) is the formula, do the math from there and you get "N-3"
The first five terms of the sequence defined by (4n) can be found by substituting (n) with the integers 1 through 5. Thus, the terms are: For (n = 1): (4 \times 1 = 4) For (n = 2): (4 \times 2 = 8) For (n = 3): (4 \times 3 = 12) For (n = 4): (4 \times 4 = 16) For (n = 5): (4 \times 5 = 20) So, the first five terms are 4, 8, 12, 16, and 20.
The number that is 4 times larger than ( n ) can be expressed as ( n + 4n ), which simplifies to ( 5n ). Therefore, if you want to determine a quantity that is 4 times larger than ( n ), you would calculate ( 5n ).
4*(n + 1) = 6*n*3
9
The question is open to multiple interpretations but I think you mean [(-2m)^4] x (n^6)^2 = [(-2)^4](m^4)(n^12) = 16(m^4)(n^12) or 16 times m to the 4th power times n to the 12th power.
40
16(2^n)(10)(2^n)=160[2^(2n)]=160(4^n)
n = number. 47 * n + 4 or 47(n + 4)
To find the number of different groups of 4 that can be made from 17 students, you can use the combination formula, which is given by ( C(n, r) = \frac{n!}{r!(n-r)!} ). In this case, ( n = 17 ) and ( r = 4 ). Therefore, the calculation is ( C(17, 4) = \frac{17!}{4!(17-4)!} = \frac{17!}{4! \times 13!} = \frac{17 \times 16 \times 15 \times 14}{4 \times 3 \times 2 \times 1} = 2380 ). Thus, there are 2,380 different groups of 4 that can be formed from 17 students.
10n = -30 n = -3
It is 4n
(n+4)2