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Masteriza [31]
3 years ago
7

Wren’s first display box is 6 inches long, 9 inches wide, and 4 inches high. What is the volume of the display box? Explain your

work using a diagram.

Mathematics
1 answer:
Black_prince [1.1K]3 years ago
7 0

Answer:

\text{Volume of display box}=216\text{ in}^3    

Step-by-step explanation:

We have been given that Wren’s first display box is 6 inches long, 9 inches wide, and 4 inches high. We are asked to find the volume of display box.

We know that the box is in form of cuboid, so its volume would be length times width times height.

\text{Volume of display box}=6\text{ in}\times 9\text{ in}\times 4\text{ in}

\text{Volume of display box}=216\text{ in}^3

Therefore, the volume of the display box is 216 cubic inches.

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Which expression is equivalent to...? Assume... screenshots attached, please help!
larisa [96]

Answer:

\frac{\sqrt[3]{5} }{3x}

Step-by-step explanation:

Ok, let's do this step by step....

\sqrt[3]{\frac{10x^{5} }{54x^{8} } }

Let's first simplify the x's:

\sqrt[3]{\frac{10}{54x^{3} } }

Then we breakdown the 54 as 2 * 27 then simplify with the 10 above.

\sqrt[3]{\frac{10}{2 * 27x^{3} } } = \sqrt[3]{\frac{5}{27x^{3} } }

Now, we can rewrite this as the following and solve the bottom part:

\frac{\sqrt[3]{5} }{\sqrt[3]{27x^{3} } } = \frac{\sqrt[3]{5} }{3 \sqrt[3]{x^{3} } } = \frac{\sqrt[3]{5} }{3x}

The solution is

\frac{\sqrt[3]{5} }{3x}

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4 years ago
When constructing an inscribed polygon with a compass and a straightedge, how should I start the construction?
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There are 50 students in the 5th grade at Clark School. One day, 20% of them were absent. How many fifth graders were in school
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Step-by-step explanation:

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