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Readme [11.4K]
2 years ago
14

Someone help me please :)

Mathematics
1 answer:
Montano1993 [528]2 years ago
5 0

Answer:

sorry kid idontknow

Step-by-step explanation:

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ILL GIVE BRAINLIEST!!!!!!<br><br> Please explain!! Thank youuuuu!
Natali5045456 [20]

Answer:

Table D

Step-by-step explanation:

you have to solve .3x = 1.5. this equals 5....since 5 is doubled to 10 you have to double 1.5 to 3.0

5 0
3 years ago
Read 2 more answers
A man gets an invoice for ​$460 with terms 4/10, ​1/15, n/30. How much would he pay 6 days after the invoice​ date?
7nadin3 [17]

Answer:

$441.60

Step-by-step explanation:

Invoice amount: $460

Term 4/10= 4% discount if paid early within the first 10 days of the invoice date

Amount due: $441.60

Discount amount: $18.40

Invoice amount: $460

Term: 1/15= 1% discount if paid early within the first 15 days of the invoice date

Amount due: $455.40

Discount amount: $4.60

Invoice amount: $460

Term: n/30= no discount if paid on or after the 30 days from the invoice date

Amount due: $460

Discount amount: $0

3 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
Your answer<br> What is the greatest common factor of 12 and 24?<br> Your answer<br> 6
anastassius [24]

Answer:

yes it is 6

Step-by-step explanation:

Both are even numbers and caan be divided by 6 win 24 is divided by two I equals 12.

8 0
3 years ago
The length of a rectangle is shown below: On a coordinate grid from negative 6 to positive 6 on the x-axis and on the y-axis two
Degger [83]

9514 1404 393

Answer:

  3.  C(−1, −5), D(2, −5)

Step-by-step explanation:

The given points are 2 -(-1) = 3 units apart horizontally, so the remaining points will need to be 18/3 = 6 units below, vertically. Translating the given points down 6 gives ...

  D = A -(0, 6) = (2, 1) -(0, 6) = (2, -5)

  C = B -(0, 6) = (-1, 1) -(0, 6) = (-1, -5)

The two points required to make a rectangle with an area of 18 square units are ...

  C(-1, -5), D(2, -5)

_____

The area of a rectangle is the product of length and width. If the area is 18 square units, and the width is 3 units, then the length must be 6 units. 3×6=18.

5 0
2 years ago
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