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The expression (4n^2) represents a quadratic function where (n) is a variable. To find the first five terms, we can substitute (n) with the integers 1 through 5:
- For (n=1): (4(1^2) = 4)
- For (n=2): (4(2^2) = 16)
- For (n=3): (4(3^2) = 36)
- For (n=4): (4(4^2) = 64)
- For (n=5): (4(5^2) = 100)
Thus, the first five terms are 4, 16, 36, 64, and 100.
What else can I help you with?
What is the square root of 4n2?
sqrt(4n2) = 2n
What is the GCF of 12n2 and 16n4?
4n2
What is the answer to 4n2-15n-25 equals?
when n = 0 it equals -25 when n = 1 it equals -36 when n = 2 it equals -39 etc
Is the square of an odd integer odd?
Yes. Any odd integer, k, is of the form 2n+1 where n is an integer. then k2 = (2n + 1)2 = 4n2 + 4n + 1. The first two terms are even so the three-term total must be odd.
What are the first 5 terms of n plus 3?
The expression "n plus 3" can be represented as ( n + 3 ). To find the first five terms, we can substitute the values ( n = 1, 2, 3, 4, ) and ( 5 ) into the expression. The first five terms are: ( 1 + 3 = 4 ) ( 2 + 3 = 5 ) ( 3 + 3 = 6 ) ( 4 + 3 = 7 ) ( 5 + 3 = 8 ) Thus, the first five terms are 4, 5, 6, 7, and 8.
How do i do 5 plus 5mn- 11mn?
8n + 4n2 -8n
What is the square root of 4n2?
sqrt(4n2) = 2n
2mn plus 4n2 and 5m2 - 7mn?
9m2+ 9mn- 4n2
What is the answer for this problem 2mn 4n2 and 5m2 - 7mn?
2mn + 4n2 and 5m2 - 7mn
What is the sum of this polynomial 2mn plus 4n2 and 5m2 - 7mn?
4n2 + 5m2 - 5mn
What is the greatest common factor of the pair of expressions 4n2?
4n2 is one expression. You need at least two to find a GCF
How do you solve 4n2 62n-32?
If you mean: 4n^2+62n-32 then it is (2n-1)(n+16) when factored after dividing all terms by 2
What is the GCF of 12n2 and 16n4?
4n2
How do you solve 4n squared-25 equals 0?
4n2-25 = 0 4n2=25 n2=6.25 n=sqrt(6.25) n= +2.5 or -2.5
What is the answer to 4n2-15n-25 equals?
when n = 0 it equals -25 when n = 1 it equals -36 when n = 2 it equals -39 etc
Can you prove If t is an odd multiple of 5 then t2 will be anumber that ends with 25?
It t is an odd multiple of 5, you can express it as 5(2n+1), where "n" is some integer. The square of that expression is 25(2n+1)2 = 25(4n2+4n+1) = 100n2+100n+25. Since the first two terms are multiples of 100 (remember, "n" is an integer), you have some multiple of 100, plus 25 - in other words, a number that ends with 25.
Is the square of an odd integer odd?
Yes. Any odd integer, k, is of the form 2n+1 where n is an integer. then k2 = (2n + 1)2 = 4n2 + 4n + 1. The first two terms are even so the three-term total must be odd.
