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4 years ago

8

What is the smallest edge possible for a cube that will hold 1 litre of water? Remember: 1cm3 = 1mL

1 answer:

Bas_tet [7]4 years ago

3 0

Answer: 10 mml long

Step-by-step explanation:

The volume of liquids is usually measured in liters or milliliters. 1 liter = 1000 ml and 1 ml = 1 cm 3. 1 liter ... 1 cm. 1 cm. 1 cubic centimeter. 1 cm 3. 3 cm. 2 cm. 4 cm. 1 cubic millimeter. 1 mm. 1 mm. 1 mm ... How much sand will it take to fill the sandpit to a depth of 20 centimeters? ... 2 All the edges of this dice are 10 mm long.

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Any help is appreciated!!

qaws [65]

Answer:

Step-by-step explanation:

600,000,000

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3 years ago

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NEED HELP ASAP!!!! <br> A) 12 sq. cm.<br> B) 6 sq. cm.<br> C) 12 cm<br> D) 6 cm

zaharov [31]

Answer:

its B- 6 sq. cm.

Step-by-step explanation:

1/2 times base times height

1/2 x 4 x 3

6

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2 years ago

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Translate the sentence into an equation.

Vlad1618 [11]

Answer:

3w+7=9

Step-by-step explanation:

i need at least 20 characters so I just rote this

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4 years ago

torisob [31]

Answer:

666 divided by 6 is 111. So, they would need 111 vans.

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3 years ago

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After being treated with chemotherapy, the radius of the tumor decreased by 23%. What is the corresponding percentage decrease i

Crazy boy [7]

Answer:

The volume of the tumor experimented a decrease of 54.34 percent.

Step-by-step explanation:

Let suppose that tumor has an spherical geometry, whose volume ( $V$ ) is calculated by:

$V = \frac{4\pi}{3}\cdot R^{3}$

Where $R$ is the radius of the tumor.

The percentage decrease in the volume of the tumor ( $\%V$ ) is expressed by:

$\%V = \frac{\Delta V}{V_{o}} \times 100\,\%$

Where:

$\Delta V$ - Absolute decrease in the volume of the tumor.

$V_{o}$ - Initial volume of the tumor.

The absolute decrease in the volume of the tumor is:

$\Delta V = V_{o}-V_{f}$

$\Delta V = \frac{4\pi}{3}\cdot (R_{f}^{3}-R_{o}^{3})$

The percentage decrease is finally simplified:

$\%V = \left[1-\left(\frac{R_{f}}{R_{o}}\right)^{3} \right]\times 100\,\%$

Given that $R_{o} = R$ and $R_{f} = 0.77\cdot R$ , the percentage decrease in the volume of tumor is:

$\%V = \left[1-\left(\frac{0.77\cdot R}{R}\right)^{3} \right]\times 100\,\%$

$\%V = 54.34\,\%$

The volume of the tumor experimented a decrease of 54.34 percent.

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3 years ago