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A right rectangular prism has a height 9 cm. The length of the base is x cm and it is 7 cm more than the width of the base. Write a function of the volume (V) of the prism in terms of the length of the base.

Mathematics

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Answers

2 parts
3

D\text{D}

4

A\text{A}

Step-by-step solution

8 steps

Part 3· 4 steps

  1. Identify the given dimensions of the right rectangular prism.

    The height h=9h = 9 cm.
    The length of the base l=xl = x cm.
    The length is 7 cm more than the width, so l=w+7l = w + 7.

  2. Express the width in terms of xx.

    Since l=w+7l = w + 7 and l=xl = x, we have x=w+7x = w + 7.
    Solving for ww:

    w=x−7w = x - 7
  3. Write the formula for the volume of a rectangular prism.

    The volume VV is given by the formula V=length×width×heightV = \text{length} \times \text{width} \times \text{height}.

    V=l×w×hV = l \times w \times h
  4. Substitute the expressions for length, width, and height into the volume formula.

    V(x)=x×(x−7)×9V(x) = x \times (x - 7) \times 9 V(x)=9x(x−7)V(x) = 9x(x - 7)

    Comparing this with the given options, the correct function is D.
    The final answer is D\boxed{\text{D}}.

Part 4

A\text{A}

  1. Split the compound inequality into two separate inequalities.

    The given inequality is 2x−3≤4x+5<3x+72x - 3 \leq 4x + 5 < 3x + 7.
    This can be written as two inequalities:

    1. 2x−3≤4x+52x - 3 \leq 4x + 5
    2. 4x+5<3x+74x + 5 < 3x + 7
  2. Solve the first inequality.

    2x−3≤4x+52x - 3 \leq 4x + 5

    Subtract 2x2x from both sides:

    −3≤2x+5-3 \leq 2x + 5

    Subtract 5 from both sides:

    −3−5≤2x-3 - 5 \leq 2x −8≤2x-8 \leq 2x

    Divide by 2:

    −82≤2x2\frac{-8}{2} \leq \frac{2x}{2} −4≤x-4 \leq x
  3. Solve the second inequality.

    4x+5<3x+74x + 5 < 3x + 7

    Subtract 3x3x from both sides:

    x+5<7x + 5 < 7

    Subtract 5 from both sides:

    x<7−5x < 7 - 5 x<2x < 2
  4. Combine the solutions from both inequalities.

    We have x≥−4x \geq -4 and x<2x < 2.
    Combining these, the solution is:

    −4≤x<2-4 \leq x < 2

    Comparing this with the given options, the correct solution is A.
    The final answer is A\boxed{\text{A}}.

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