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5n (also known as 5xn)=10, which means n=2. n is the variable. 5-n=2 which means n=3 5/n (I'm trying to say 5 over n, also known as 5 divided by n)=20, which means n=0.25. 5+n=7 n=2.
n + 2 = 5, solve for n.n = 5 - 2n = 3
Formula: number of diagonals = 1/2 * n(n-3) where n is the number of sides or vertices Example: to get the number of diagonals in a pentagon pentagon has 5 sides or vertices; thus, n = 5 no. of diagonals: = 1/2 * n(n-3) = 1/2 * 5(5-3) = 1/2 * 5(2) = 1/2 * 10 = 10/2 = 5
Let number be N, then 2 x N + 5 = 45 2 x N = 45 - 5 = 40 N = 20
The sequence of numbers is 2n where n goes from 1 to 5 Now sum (over n=1 to 5) of 2n = 2*sum (over n=1 to 5) of n = 2 * [5*(5+1)]/2 = 5*6 = 30
That would be 18+n. 5+3+2 = 5+3+n 10 = 8 + n 10-8 = n 2 = n and if you substitute 2 for n then you see that 5+3+2 = 5+3+2
Yes. A polygon has n(n-3)/2 diagonals, so if n=5: 5(2)/2 =5
n ___ = 3 2+5 n = 3(2+5) n = 3(7) n = 21
If we are to find the product of 5 and m and n/2 (which is half of n), we have: 5 times m times n/2 = 5 x m x n/2 = 5mn/2
5n (also known as 5xn)=10, which means n=2. n is the variable. 5-n=2 which means n=3 5/n (I'm trying to say 5 over n, also known as 5 divided by n)=20, which means n=0.25. 5+n=7 n=2.
The value of the expression n(n-1)(n-2)(n-3)(n-4)(n-5) is the product of n, n-1, n-2, n-3, n-4, and n-5.
n + 2 = 5, solve for n.n = 5 - 2n = 3
Formula: number of diagonals = 1/2 * n(n-3) where n is the number of sides or vertices Example: to get the number of diagonals in a pentagon pentagon has 5 sides or vertices; thus, n = 5 no. of diagonals: = 1/2 * n(n-3) = 1/2 * 5(5-3) = 1/2 * 5(2) = 1/2 * 10 = 10/2 = 5
5-2=3n=3
Let's break this down step by step. If we add 5 to the number n, we get n + 5. When we multiply this sum by 2, we get 2(n + 5). Given that this result is equal to 24, we can set up the equation 2(n + 5) = 24. Solving for n, we find n = 7.
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Let number be N, then 2 x N + 5 = 45 2 x N = 45 - 5 = 40 N = 20