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Nat2105 [25]
2 years ago
8

What inequality is graphed on the number line below?

Mathematics
2 answers:
vesna_86 [32]2 years ago
5 0

ANSWER: The inequality graphed on the number line is: C. b <u><</u> 3

vazorg [7]2 years ago
4 0

Answer:

  • C) b ≤ 3

Step-by-step explanation:

The endpoint of the graph is the midpoint between 2 and 4, so it is 3.

This point is included since indicated as full circle.

<u>The graph includes the line to the left of 3:</u>

  • b ≤ 3 is the correct one
You might be interested in
Find the value of x.
maxonik [38]

180 = 132 + x + x

180 - 132 = 2x

48 = 2x

x = 24⁰

5 0
2 years ago
Read 2 more answers
Can someone help me with these and show the work also?
Snezhnost [94]

QUESTION:

Simplify each expression

ANSWER:

1.) \green{{- 8n}}

2.) \green{{- 2b - 60}}

3.) \green{{- 10x - 14}}

4.) for number 4 study my step-by-step explanation so you can answer that

STEP-BY-STEP EXPLANATION:

1.) First, If the term doesn't have a coefficients, it is considered that the coefficients is 1

WHY?

Learn why:

Why is it considered that the coefficient is 1?

Remember that any term multiplied by \blue{{1}} remains the same :

\blue{{1}} {× x = x}

Step 1:

The equality can be read in the other way as a well, so any term can be written as a product of \blue{{1}} and itself:

{x = } \blue{{1}} {× x}

Step 2:

Usually, we don't need to write multiplacation sign between the coefficient and variable, so the simple form is:

{x = 1x}

This is why we can write the term without the coefficient as a term with coefficient {1}

Now let's go back to solving as what i said if a term doesn't have a coefficient, it is considered that the coefficient is 1

{n - 9n}

\red{{1}} {n -9n}

Second, Collect like terms by subtracting their coefficients

\red{{1n - 9n}}

\red{{( 1 - 9)n}}

Third, Calculate the difference

how?

Keep the sign of the number with the larger absolute value and subtract the smaller absolute value from larger

\red{{1 - 9}}

\red{{- (9 - 1)}}

Subtract the numbers

- (\red{{9 - 1}})n

- \red{{8}}n

\green{\boxed{- 8n}}

2.) First, Distribute - 6 through the parentheses

how?

Multiply each term in the parentheses by - 6

\red{{- 6(b + 10)}}

\red{{- 6b - 6 × 10}}

Multiply the numbers

- {6b} - \red{{6 × 10}}

- {6b} - \red{{60}}

Second, Collect like term

how?

Collect like terms by calculating the sum or difference of their coefficient

\red{{- 6b + 4b}}

\red{{(- 6 + 4)b}}

Calculate the sum

\red{{(- 6 + 4)}}b

\red{{-2}}b

\green{\boxed{- 2b - 60}}

3.) First, Distribute 2 through parentheses

how?

Multiply each term in the parentheses by 2

\red{{2(x - 5)}}

\red{{2x - 2 × 5}}

Multiply the numbers

{2x -} \red{{2 × 5}}

{2x -} \red{{10}}

Second, Distribute - 4 through the parentheses

how?

Multiply each term in the parentheses by - 4

\red{{- 4(3x + 1)}}

\red{{- 4 × 3x - 4}}

Calculate the product

- \red{{4 × 3}}x - 4

- \red{{12}}x - 4

Third, Collect like terms

how?

Collect like terms by subtracting their coefficient

\red{{2x - 12x}}

\red{{(2 - 12)x}}

Calculate the difference

\red{{(2 - 12)}}x

\red{{- 10}}x

Fourth, Calculate the difference

how?

Factor out the negative sign from the expression

\red{{- 10 - 4}}

\red{{- (10 + 4)}}

Add the numbers

- (\red{{10 + 4}})

- \red{{14}}

\green{\boxed{- 10x - 14}}

That's all I know sorry but I hope it helps :)

6 0
2 years ago
Which two comparisons are true?
uranmaximum [27]

Answer:

The answer is b and d

...........

8 0
2 years ago
Arrange the geometric series from least to greatest based on the value of their sums.
son4ous [18]

Answer:

80 < 93 < 121 < 127

Step-by-step explanation:

For a geometric series,

\sum_{t=1}^{n}a(r)^{t-1}

Formula to be used,

Sum of t terms of a geometric series = \frac{a(r^t-1)}{r-1}

Here t = number of terms

a = first term

r = common ratio

1). \sum_{t=1}^{5}3(2)^{t-1}

   First term of this series 'a' = 3

   Common ratio 'r' = 2

   Number of terms 't' = 5

   Therefore, sum of 5 terms of the series = \frac{3(2^5-1)}{(2-1)}

                                                                      = 93

2). \sum_{t=1}^{7}(2)^{t-1}

   First term 'a' = 1

   Common ratio 'r' = 2

   Number of terms 't' = 7

   Sum of 7 terms of this series = \frac{1(2^7-1)}{(2-1)}

                                                    = 127

3). \sum_{t=1}^{5}(3)^{t-1}

    First term 'a' = 1

    Common ratio 'r' = 3

    Number of terms 't' = 5

   Therefore, sum of 5 terms = \frac{1(3^5-1)}{3-1}

                                                 = 121

4). \sum_{t=1}^{4}2(3)^{t-1}

    First term 'a' = 2

    Common ratio 'r' = 3

    Number of terms 't' = 4

    Therefore, sum of 4 terms of the series = \frac{2(3^4-1)}{3-1}

                                                                       = 80

    80 < 93 < 121 < 127 will be the answer.

4 0
2 years ago
Read 2 more answers
(1/3+9) x ( 8-3) tell me the steps help
mafiozo [28]
Use PEMDAS to solve
(Parenthesis, Exponents, Multiplication, Division, Addition, Subtraction)

Solve parenthesis first
<span>(1/3+9) x ( 8-3)
</span>28/3 x 5

Use Multiplication 
28/3 x 5 ≈ 4.7
After rounding 4.7 should be your answer
3 0
3 years ago
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