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What are the first 4 mutpules of 2n?
I am assuming that 2n is an algebraic expression, and n is limited to positive integer values. The first 4 multiples of 2n are 0 (2n*0), 2n (2n*1), 4n (2n*2), and 6n (2n*3). If you are looking for non-0 multiples, you would also include 8n (2n*4).
How do you integrate sin x to the power of 2n?
I will assume that we are taking d/dx, not d/dn. There are two ways to interpret what you asked. First way is (sinx)^(2n). Second way is sin(x^(2n)). First answer: 2n(sinx)^(2n-1)(cosx)=2ncosx(sinx)^(2n-1). Second answer: cos(x^(2n))(2nx^(2n-1)).
What is the first 3 terms if the nth term is 2n?
n = 1, 2n = 2 n = 2, 2n = 4 n = 3, 2n = 6 2, 4, 6, ..., 2n where n = 1, 2, 3, ... This is an arithmetic sequence, where the first term is 2 and the common difference is 2.
Twice the first number is 6 less than 2 times the first?
This question does not make sense...You are saying that 2n - 6 = 2n, which is impossible
What is the first four multiples of 2n?
2,4,6, and 8
What is 2n-3 equals 9?
2n-3 = 9 solve for n by first adding 3 to both sides 2n = 12 divided both sides by 2 n = 6
What does 2N and 2N mean?
2n+2n equals 4n
Find the first 3 terms and 50th and 100th term for 2n plus 4?
2n+4: 6,8,10......104........204
How do you pick out the factors of xy2n2 x2y2n?
To factor the expression ( xy^2n^2 x^2y^2n ), first, identify the common factors in each term. The expression can be rewritten as ( (xy^2n^2)(x^2y^2n) ). The common factors include ( xy^2n ), which appears in both terms. Thus, the factors can be expressed as ( xy^2n (xn^2 + x^2y) ).
What are three consecutive even integers whose sum is 108?
Let 2n be the first integer, then the next on is 2n+2 and the one after is 2n+4 so adding all three we have 6n+6=108 or 6n=102 and we want 2n so divide we divide 6n by 3 and we have 2n=34, and 2n+2=36 and lastly 2n+4=38 so 34, 36, 38 are the even integers we seek.
What is (-1)(2n)(4n)?
FOIL deals with polynumerals in math. An example is a difference of squares as in (2n - 2)2 or (2n + 2)(2n - 2) =F stand for first--2n X 2n = 4n2O stands for outside-- 2n X -2 = -4nI stands for inside-- -2 X 2n = 4nL stands for last--2 X -2 = -4In the end you get:4n2 - 4n + 4n - 4 = 4n2 - 4Can you see how everything is squared and the two numbers are being subtracted?
What are the first 3 terms for 2n-6?
The first three terms for the expression 2n-6 are obtained by substituting n with consecutive integers. When n=1, the expression evaluates to -4; when n=2, the expression evaluates to -2; and when n=3, the expression evaluates to 0. Therefore, the first three terms are -4, -2, and 0.
