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Aleksandr-060686 [28]
3 years ago
15

Which expression represents the fourth term in the binomial expansion of (e + 2f)10?

Mathematics
2 answers:
torisob [31]3 years ago
8 0

Answer:

a.

Step-by-step explanation:

GuDViN [60]3 years ago
4 0

Answer:

first one

Step-by-step explanation:

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A line passes through (2, 8) and (4, 12). Which equation best represents the line?
andrezito [222]
If you plug in the x values of both (2, 8) and (4, 12) into each equation, you'll find that y = 2x + 4 gives you the correct y values for both (2, 8) and (4, 12). So that is your answer. 
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What is the missing length 11 + = ×16 = ÷2 =209
zzz [600]
11 plus what number equals 16 in order to complete your problem.
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Pls help...............
Alex Ar [27]

Answer:

You chose the right one

Step-by-step explanation:

i hope this helps :)

8 0
2 years ago
If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar
sleet_krkn [62]

This question is incomplete, the complete question is;

If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple are written in increasing order but are not necessarily distinct.

In other words, how many 5-tuples of integers  ( h, i , j , m ), are there with  n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1 ?

Answer:

the number of 5-tuples of integers from 1 through n that can be formed is [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

Step-by-step explanation:

Given the data in the question;

Any quintuple ( h, i , j , m ), with n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1

this can be represented as a string of ( n-1 ) vertical bars and 5 crosses.

So the positions of the crosses will indicate which 5 integers from 1 to n are indicated in the n-tuple'

Hence, the number of such quintuple is the same as the number of strings of ( n-1 ) vertical bars and 5 crosses such as;

\left[\begin{array}{ccccc}5&+&n&-&1\\&&5\\\end{array}\right] = \left[\begin{array}{ccc}n&+&4\\&5&\\\end{array}\right]

= [( n + 4 )! ] / [ 5!( n + 4 - 5 )! ]

= [( n + 4 )!] / [ 5!( n-1 )! ]

= [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

Therefore, the number of 5-tuples of integers from 1 through n that can be formed is [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

4 0
3 years ago
What is 37.726 rounded to the nearest tenth
pochemuha
If the number you are rounding is followed by 5, 6, 7, 8, or 9, round the number up (+1).

If the number you are rounding is followed by 0, 1, 2, 3, or 4, round the number down (no change).

37.7\fbox26\approx\huge\boxed{37.7}
7 0
3 years ago
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