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NemiM [27]
3 years ago
8

Algebra 1 questions help, need help quickly. Thanks!

Mathematics
1 answer:
atroni [7]3 years ago
4 0

Answer:

6. A

7. C

8. A.

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Consider the expression 6x6^0x6^-3
Montano1993 [528]

Answer:

6*(6^-3)=0.027777 recurring

Step-by-step explanation:

6^0=1

6^-3=0.004629629 recuuring

0.004629629 recurring *6=0.0.02777recurring

6 0
3 years ago
1v1=p2v2, where p1 and v1 are the initial pressure and volume of a gas and p2 and v2 are the final pressure and volume of the ga
Dima020 [189]

Step-by-step explanation:

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7 0
3 years ago
Three times a larger number is 30 more than 5 times a smaller number. The sum of the larger number and 5 times the smaller numbe
kicyunya [14]

Answer: Smaller number = 6

Larger number = 20

Step-by-step explanation:

Let the smaller number be x.

Let the larger number be y.

Three times a larger number is 30 more than 5 times a smaller number. This will be:

3 × y = (5 × x) + 30

3y = 5x + 30 ...... i

The sum of the larger number and 5 times the smaller number is 50. This will be:

y + 5x = 50 ...... ii

From equation ii

y + 5x = 50.

y = 50 - 5x ...... iii

Put the value of y = 50 - 5x into equation i

3y = 5x + 30

3(50 - 5x) = 5x + 30

150 - 15x = 5x + 30

150 - 30 = 5x + 15x

120 = 20x

x = 120/20

x = 6

The smaller number is 6

Note that y + 5x = 50

y + 5(6) = 50

y + 30 = 50

y = 50 - 30

y = 20

The bigger number is 20

7 0
3 years ago
PLEASE HELP ASAP THANK YOU!!
Stells [14]

Answer:

Second graph

Step-by-step explanation:

See attached image.

3 0
2 years ago
Homework
Zolol [24]

<u><em>Note:</em></u><em> As you have missed to mention the first four terms of the Arithmetic sequence. So, I am randomly assuming that first four terms of the arithmetic sequence be 1, 3, 5, 7... This would anyhow make you understand the concept. So, I am solving your query based on assuming the first four terms of an Arithmetic sequence as 1, 3, 5, 7...</em>

Part A)

<em><u>What is the next term of this sequence?</u></em>

Answer:

{\displaystyle \ a_{5}=9 is the next term i.e. 5th term of the arithmetic sequence <em>1, 3, 5, 7...</em>

Step-by-step explanation:

Considering the Arithmetic sequence with fist four terms

<em> 1, 3, 5, 7...</em>

As we know that a sequence is termed as arithmetic sequence of numbers if the difference of any two consecutive terms of the sequence remains constant.

For instance, <em> 1, 3, 5, 7... </em>will be an arithmetic sequence having the common difference 2. Common difference is denoted by 'd'.

So,

Given the sequence

<em>1, 3, 5, 7...</em>

d=3-1=2,d=5-3=2

As a_{1} = 1 and d = 2

The next term i.e. 5th term can be found by using the nth term of the sequence.

So, consider the nth term of the sequence {\displaystyle a_{n}

{\displaystyle \ a_{n}=a_{1}+(n-1)d}

Putting n=5 in, a_{1} = 1 and d = 2  in {\displaystyle \ a_{n}=a_{1}+(n-1)d} to find the 5th term.

{\displaystyle \ a_{n}=a_{1}+(n-1)d}

{\displaystyle \ a_{5}=1+(5-1)2}

{\displaystyle \ a_{5}=1+(4)2}

{\displaystyle \ a_{5}=9

So, {\displaystyle \ a_{5}=9 is the next term i.e. 5th term of the arithmetic sequence <em>1, 3, 5, 7...</em>

Part B)

<u><em>Writing down an expression,  in terms of n for the nth term of the sequence</em></u>

consider the nth term of the sequence {\displaystyle a_{n}

{\displaystyle \ a_{n}=a_{1}+(n-1)d}

Here, a_{1} is the first term, d is the common difference.

For example,

Given the sequence

<em>1, 3, 5, 7...</em>

d=3-1=2,d=5-3=2

As a_{1} = 1 and d = 2

The next term i.e. 5th term can be found by using the nth term of the sequence.

So, consider the nth term of the sequence {\displaystyle a_{n}

{\displaystyle \ a_{n}=a_{1}+(n-1)d}

Putting n=5 in, a_{1} = 1 and d = 2  in {\displaystyle \ a_{n}=a_{1}+(n-1)d} to find the 5th term.

{\displaystyle \ a_{n}=a_{1}+(n-1)d}

{\displaystyle \ a_{5}=1+(5-1)2}

{\displaystyle \ a_{5}=1+(4)2}

{\displaystyle \ a_{5}=9

Keywords: arithmetic sequence, nth term, common difference

Learn more abut arithmetic sequence, nth term and common difference from brainly.com/question/12227567

#learnwithBrainly

7 0
4 years ago
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