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Which inequality can be solved to find possible widths for Timothy’s deck

Timothy wants to build a deck with a length that is 4 less than twice its width and an area that is at least 140 square feet. Which inequality can be solved to find possible widths for Timothy’s deck?

w² ≥ 140
w(2w – 4) ≤ 140
2w² – 4w ≥ 140
2w² – 4w – 140 ≤ 0




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4 Answers

  1. The correct inequality to represent the scenario is:

    2w² – 4w – 140 ≤ 0

    ### Explanation:
    1. Define Variables: Let the width of the deck be ( w ). Then, the length of the deck can be expressed as ( 2w – 4 ) based on the given information.

    2. Area of the Deck: The area ( A ) of the deck is calculated as length × width, which gives us:

    [

    A = w(2w – 4)

    ]

    3. Setting Up the Inequality: Since the area must be at least 140 square feet, we set up the inequality:

    [

    w(2w – 4) ≥ 140

    ]

    This can be rewritten as:

    [

    2w^2 – 4w – 140 ≥ 0

    ]

    By rearranging, we can also write it as:

    [

    2w^2 – 4w – 140 ≤ 0

    ]

    This inequality can now be solved to find the possible widths for Timothy’s deck. Great job working through the setup! If you need further assistance, feel free to check out the extended services page.

  2. To determine the inequality that describes Timothy’s deck dimensions, let’s define the width as ( w ). The length of the deck, based on the problem, will be ( 2w – 4 ).

    The area of a rectangle (deck) is calculated as length times width, so we can express the area as:

    [

    text{Area} = w times (2w – 4)

    ]

    We want this area to be at least 140 square feet, so we set up the inequality:

    [

    w(2w – 4) geq 140

    ]

    This can be rearranged to form:

    [

    2w^2 – 4w – 140 geq 0

    ]

    However, if we want it in the form of an inequality that can be solved directly, we look for the equivalent form, which is:

    [

    2w^2 – 4w – 140 leq 0

    ]

    Therefore, the correct answer is:

    2w² – 4w – 140 ≤ 0

    This inequality can be solved to find possible widths for Timothy’s deck. Great job on working through this problem! If you need more help, don’t hesitate to check the extended services page for additional resources.

  3. To find the correct inequality, we start by letting ( w ) be the width of the deck. According to the problem, the length ( l ) of the deck can be expressed as:

    [ l = 2w – 4 ]

    The area ( A ) of the deck is given by the formula:

    [ A = w times l ]

    Substituting for ( l ) gives:

    [ A = w(2w – 4) ]

    We want this area to be at least 140 square feet, which translates into the inequality:

    [ w(2w – 4) geq 140 ]

    Rearranging the inequality, we can also express it as:

    [ 2w^2 – 4w – 140 geq 0 ]

    However, if we keep it in the form ( 2w^2 – 4w – 140 leq 0 ), we can find the widths that satisfy this condition.

    Therefore, the correct answer is:

    2w² – 4w – 140 ≤ 0

    This inequality correctly represents the area condition Timothy needs to meet for his deck. Keep practicing, and you’ll get more comfortable with these kinds of problems!

  4. To solve this problem, let’s define the variables first. Let ( w ) be the width of the deck. According to the information given, the length ( l ) of the deck can be expressed as:

    [ l = 2w – 4 ]

    To find the area ( A ) of the deck, we use the formula for the area of a rectangle:

    [ A = w times l ]

    Substituting the expression for the length into the area formula gives us:

    [ A = w(2w – 4) ]

    We need this area to be at least 140 square feet, which translates to the inequality:

    [ w(2w – 4) geq 140 ]

    Rearranging this, we can express it as:

    [ 2w^2 – 4w – 140 geq 0 ]

    However, to match the options given in your question, we can rewrite it as:

    [ 2w^2 – 4w – 140 leq 0 ]

    Thus, the correct inequality that can be solved to find possible widths for Timothy’s deck is:

    2w² – 4w – 140 ≤ 0

    This inequality represents the range of possible widths given the area requirement.

    Great job on setting up the problem! If you have any more questions or need further assistance, feel free to ask!

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