Question 6 Answered

Question 6 (1 point)
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Which expression is equivalent to
4-2(5-x)
Question 6 options:

-2x-6

2x-6

7-2x

10-2x

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Question 10 Needs Answer
Question 10 (1 point)
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Which equation is a result of the first step in solving the equation below?

Question 10 options:

Question 12 Needs Answer
Question 12 (1 point)
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Find the value of h in the equation below

-4(5h+7)+16h=40
Question 12 options:

h = -2

h = -17

h = -3

h = 17

Question 13 Needs Answer
Question 13 (1 point)
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The equation below was solved incorrectly. Describe the error and solve correctly for x.

Question 13 options:

x = 1; They did not correctly distribute the -5 to both terms in (x+1)

x = -1; They did not correctly combine x-5x

x = 1; They did not correctly combine x - 5x

x = -1; They did not correctly distribute the -5 to both terms in (x+1)

Question 14 Needs Answer
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Find the value of a in the equation below

16-5a+2a-1=41-a
Question 14 options:

a = -28

a = 13

a = -13

a = 28

Question 16 Needs Answer
Question 16 (1 point)
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Consider the equation

What can you replace the ? with to create an equation with INFINITE SOLUTIONS?
2(4-a)=
Question 16 options:

-4

4

2

1

Question 18 Needs Answer
Question 18 (1 point)
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Find the value of p in the equation below

-13+12p-4=6(2p-1)
Question 18 options:

p = -17

p = 2

No Solution

Infinite Solutions

Question 19 Needs Answer
Question 19 (1 point)
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Ella is selling a dozen cookies for $4. She had to pay $20 for ingredients. How much will she earn if she sells 12 dozen cookies?

Question 19 options:

$48

$28

$68

$288

1 answer

Let's go through each question that you mentioned and provide the answers.

Question 6: Which expression is equivalent to 4 - 2(5 - x)?

To solve this, distribute the -2:

\[ 4 - 2(5 - x) = 4 - 10 + 2x = -6 + 2x = 2x - 6 \]

So the correct option is: 2x - 6


Question 12: Find the value of h in the equation -4(5h + 7) + 16h = 40

First, distribute -4:

\[ -20h - 28 + 16h = 40 \]

Combine like terms:

\[ -4h - 28 = 40 \]

Add 28 to both sides:

\[ -4h = 68 \]

Divide by -4:

\[ h = -17 \]

So the correct answer is: h = -17


Question 13: Describe the error and solve correctly for x.

Without the specific equation in question, I can only provide you with a general method. It seems like the options you provided are stating likely errors. In the case of distributing incorrectly, here's what you would check:

If the equation was \( -5(x + 1) = ax + b \), the correct distribution would be \( -5x - 5 \). Check if that was done correctly, and then proceed to solve for \( x \).

Based on the choices, if they did not distribute correctly, then: x = 1; They did not correctly distribute the -5 to both terms in (x + 1) would likely be correct if that matches the context.


Question 14: Find the value of a in the equation 16 - 5a + 2a - 1 = 41 - a

Combine like terms:

\[ 15 - 3a = 41 - a \]

Add \( a \) to both sides:

\[ 15 - 2a = 41 \]

Subtract 15 from both sides:

\[ -2a = 26 \]

Divide by -2:

\[ a = -13 \]

So the correct answer is: a = -13


Question 16: What can you replace the ? with for infinite solutions?

For an equation to have infinite solutions, both sides must be equal in a way that any \( a \) would make the equation true. The expression you provided is \( 2(4 - a) = ? \). The left side is \( 8 - 2a \).

To make both sides equal, you could just substitute \( 8 - 2a \) for ?.

So the correct option is: 4 since both are offset by a factor that doesn’t affect the equality.


Question 18: Find the value of p in the equation -13 + 12p - 4 = 6(2p - 1)

First, simplify:

\[ -17 + 12p = 12p - 6 \]

Subtract \( 12p \) from both sides:

\[ -17 = -6 \]

This is impossible, so there is no solution.

So, the answer is: No Solution


Question 19: Ella is selling a dozen cookies for $4. She had to pay $20 for ingredients. How much will she earn if she sells 12 dozen cookies?

First, calculate the revenue from selling 12 dozen cookies:

Price per dozen = $4, hence for 12 dozen: \[ 12 \times 4 = 48 \]

Now subtract the cost of ingredients: \[ $48 - $20 = $28 \]

So the answer is: $28

If you have any more questions or need further assistance, feel free to ask!

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