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Archy [21]
2 years ago
13

Help here please. thank you

Mathematics
1 answer:
NISA [10]2 years ago
4 0

Answer: A=C+7

Step-by-step explanation:

Let’s say hypothetically C=6, then B would be 4 since you would have to subtract 2 and A would be 13 since 4+9=13

The difference between 13 and 6 is 7, but that would work with any number and i’m just using c=6 as an example to make the problem easier to understand

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Korolek [52]

Answer:

(1/2) / (1/12)

(1/2) * (12/1) = 12/2 = 6


6 0
3 years ago
Write 800 as the product of its prime factors.
Alex_Xolod [135]
800|2
800|2
400|2
200|2
100|2
50|2
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1

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8 0
3 years ago
please help im having such a hard time :( does this equation have no solution one solution or infinite solutions
WINSTONCH [101]

One solution!

Step-by-step explanation:

If we make a graph of the given equations, the lines will intersect at a point. We know that whenever lines intersect, it means the equations have one solution or a unique solution.

8 0
3 years ago
Read 2 more answers
Yasha is going to invest $3500 for six years. She is going to invest in a savings account
amm1812

Answer:

umm let me see

Step-by-step explanation:

5 0
3 years ago
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
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