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hram777 [196]
3 years ago
9

2/3 divided by 1/6 PLEASE HELP MEEE

Mathematics
2 answers:
Alex73 [517]3 years ago
7 0

Answer:

it's 4

Step-by-step explanation:

DiKsa [7]3 years ago
3 0
The answer is 4.

2/3 is equal to 4/6
4/6 divided by 1/6 is 4
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What is the function rule represented by the following table?
Romashka-Z-Leto [24]

Answer:

y=-4x+3

Step-by-step explanation:

as x=0

put in y= -4x+3

y= -4(0)+3

y=3

now put x=1 in y= -4x+3

y= -4(1)+3

y= -1

now put x=2 in y= -4x+3

y= -4(2)+3

y=-5

3 0
4 years ago
Hey guys my other account has been banned for 24 hours like wow but it ok im giving away 48 points to the first two people that
Vaselesa [24]

Answer:

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Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
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
Please answer CORRECTLY !!!!!! Will mark brainliest !!!!!!!!!
gulaghasi [49]

Answer:

a= 13+3b

Step-by-step explanation:

Simplifying

a + -7 = 3(b + 2)

Reorder the terms:

-7 + a = 3(b + 2)

Reorder the terms:

-7 + a = 3(2 + b)

-7 + a = (2 * 3 + b * 3)

-7 + a = (6 + 3b)

Solving

-7 + a = 6 + 3b

Solving for variable 'a'.

Move all terms containing a to the left, all other terms to the right.

Add '7' to each side of the equation.

-7 + 7 + a = 6 + 7 + 3b

Combine like terms: -7 + 7 = 0

0 + a = 6 + 7 + 3b

a = 6 + 7 + 3b

Combine like terms: 6 + 7 = 13

a = 13 + 3b

Simplifying

a = 13 + 3b

5 0
4 years ago
1. Which equation does not match the diagram? A. 6 +3x= 30
GalinKa [24]

Answer:

B

Step-by-step explanation:

6 0
3 years ago
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