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NNADVOKAT [17]
3 years ago
15

What is 8.023 in expanded form

Mathematics
2 answers:
aleksklad [387]3 years ago
7 0

Answer:

8 + 0.0 + 0.02+ 0.003

Step-by-step explanation:

Wewaii [24]3 years ago
4 0

Answer:

8+0.02+0.003

hope this helps

have a good day :)

Step-by-step explanation:

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Stella owns 120,000 acres of land. She plans to divide 30% of the land to be farm land. How much land does she have that is NOT
valentina_108 [34]

Answer:

40,000

Step-by-step explanation:120,000 divided by 30% is 40,00 because 120,000 divide by 30 is 40,000

8 0
4 years ago
Read 2 more answers
PLEASE HELP
alukav5142 [94]
The answer is D ……….
8 0
1 year ago
The sum of 3 odd consecutive integers are 93. find the 3 integers
Solnce55 [7]
Let, Your Integers = x, x+2, x+4
Now, x + x+2 + x+4 = 93
3x + 6 = 93
3x = 93 - 6
x = 87/3
x = 29
Then, x+2 = 31, & x+4 = 33

In short, Your Integers would be 29, 31, 33

Hope this helps!
4 0
3 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 the y-intercept of the equation of the line that is perpendicular to the line y =
igor_vitrenko [27]

Answer:

y=-5/3x+20

Step-by-step explanation:

Let the equation of the required line be represented as \[y=mx+c\]

This line is perpendicular to the line \[y=\frac{3}{5}x+10\]

\[=>m*\frac{3}{5}=-1\]

\[=>m=\frac{-5}{3}\]

So the equation of the required line becomes \[y=\frac{-5}{3}x+c\]

This line passes through the point (15.-5)

\[-5=\frac{-5}{3}*15+c\]

\[=>c=20\]

So the equation of the required line is \[y=\frac{-5}{3}x+20\]

Among the given options, option 4 is the correct one.

3 0
3 years ago
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