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Question 9
Time Remaining: 25 mins
The volume of the rectangular pyramid below is 364 units
^(3). Find the value of
x.
Answer

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Question $9$ $\newline$ Time Remaining: $25$ mins $\newline$ The volume of the rectangular pyramid below is $364$ units $^{3}$ . Find the value of $x$ . $\newline$ Answer $\newline$ $\square$ $\newline$ Submit Answer

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Solve exponential equations by rewriting the base

Full solution

Q. Question

9

\newline

Time Remaining:

25

mins

\newline

The volume of the rectangular pyramid below is

364

units

^{3}

. Find the value of

x

.

\newline

Answer

\newline

\square

\newline

Submit Answer

Understand formula for volume: Step $1$ : Understand the formula for the volume of a rectangular pyramid. $\newline$ The volume $V$ of a rectangular pyramid is given by $V = \frac{1}{3} \times \text{base area} \times \text{height}$ . $\newline$ Here, the volume is $364$ cubic units.
Identify base area and height: Step $2$ : Identify the base area and height in terms of $x$ . Assuming the base is a square with side $x$ and the height is also $x$ (since no other dimensions are provided), the base area = $x^2$ . Thus, the volume formula becomes $V = (\frac{1}{3}) \times x^2 \times x = (\frac{1}{3}) \times x^3$ .
Set up equation with volume: Step $3$ : Set up the equation with the given volume. $\newline$ $(\frac{1}{3}) \times x^3 = 364$ $\newline$ Multiply both sides by $3$ to clear the fraction: $\newline$ $x^3 = 364 \times 3$
Calculate $x^3$ : Step $4$ : Calculate $x^3$ . $\newline$ $x^3 = 1092$
Solve for x: Step $5$ : Solve for x by taking the cube root of both sides. $\newline$ $x = \sqrt[3]{1092}$ $\newline$ $x \approx 10.4$ , but let's check if this makes sense.
Verify the calculation: Step $6$ : Verify the calculation. $\newline$ $(\frac{1}{3}) \times (10.4)^3$ should be approximately $364$ . $\newline$ $(\frac{1}{3}) \times (1124.864) \approx 374.95$ , which is not $364$ . There seems to be a mistake in assuming both the base and height as $x$ .

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