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3 years ago

14

A canister contains 12 ounces of flour. When the canister is full, it holds 6 times as much flour.

2 answers:

Dimas [21]3 years ago

7 0

Answer:

72

Step-by-step explanation:

Given A canister = 12 ounces of flour. When full = 6 times more.

To find: How many ounces of flour does the canister hold when it is full?

Solution: With the given data, all that is needed to be done in order to find how many ounces can be held by the canister at its full potential, simply multiply the given data:

12 × 6 = 72

Hence, the canister can hold 72 ounces of flour when it is full.

RSB [31]3 years ago

3 0

Answer:

72

Step-by-step explanation:

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Cylinders A and B are similar. The volume of cylinder A is 20mm3. Calculate the volume of cylinder B.

MrRa [10]

The question is incomplete. The complete question is :

Cylinders A and B are similar. The length of the cylinder A is 4 mm and the length of cylinder B is 6 mm. The volume of cylinder A is 20mm3. Calculate the volume of cylinder B.

Answer:

67.5 $mm^3$

Step-by-step explanation:

Given that :

Cylinder A and cylinder B are similar.

Let volume of cylinder A = 20 $mm^3$

We know the volume of a cylinder is given by V = $\pi r^2 h$

where, r is the radius of the cylinder

h is the height of the cylinder

We have to find the scale factor.

The length scale factor is = $\frac{6}{4}$

$=\frac{3}{2}$

Area scale factor $=\left(\frac{3}{2}\right)^2$

$=\frac{9}{4}$

∴ Volume scale factor $=\left(\frac{3}{2}\right)^3$

$=\frac{27}{8}$

Therefore, the volume of cylinder B is $=20 \times \frac{27}{8}$

= 67.5 $mm^3$

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3 years ago

A rectangle measures 3 feet by 24 inches. Find the area of the rectangle in square feet.

Setler79 [48]

72 plz give brainlliest

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3 years ago

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If I roll a 6 sided dice and then flip a coin,

lina2011 [118]

Answer:

1/12

Step-by-step explanation:

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2 years ago

Nadia is investigating rotations about the center of regular polygons that carry the regular polygon onto itself. She claims tha

GalinKa [24]

Answer:

$\theta_1 \ n\ \theta_2 = 120, 240$

Step-by-step explanation:

The question is incomplete, as the angles of rotation are not stated.

However, I will list the angles less than 360 degrees that will carry the hexagon and the nonagon onto itself

We have:

$Nonagon = 9\ sides$

$Hexagon = 6\ sides$

Divide 360 degrees by the number of sides in each angle, then find the multiples.

<u>Nonagon</u>

$\theta = \frac{360}{9} =40$

List the multiples of 40

$\theta_1 = 40, 80, 120, 160, 200, 240, 280, 320$

<u>Hexagon</u>

$\theta = \frac{360}{6} =60$

List the multiples of 60

$\theta_2 = 60, 120, 180, 240, 300$

List out the common angles

$\theta_1 = 40, 80, 120, 160, 200, 240, 280, 320$

$\theta_2 = 60, 120, 180, 240, 300$

$\theta_1 \ n\ \theta_2 = 120, 240$

This means that, only a rotation of $120, 240$ will lift both shapes onto themselves, when applied to both shapes.

The other angles will only work on one of the shapes, but not both at the same time.

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3 years ago

How many miligrams are equal to 0.5 grams

Serga [27]

Answer:

500 Milligrams

4 0

2 years ago

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