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3 years ago

Whats 3/16 divided by 12

Mathematics

2 answers:

mel-nik [20]3 years ago

7 0

The answer is 0.015625

34kurt3 years ago

4 0

Answer:

0.015625

Step-by-step explana

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Please help will Mark brainiest and please show your working out thanks

AlekseyPX

Step-by-step explanation:

$800ml \times \frac{1}{4} = 200ml \\ 200ml - - > 1800ml \\ 360ml - - - > x \\ \frac{200}{360} = \frac{1800}{x} \\ x = \frac{18}{2} \times 360 \\ \: \: = 9 \times 360 \\ = 3240ml \\ = 3.24l$

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3 years ago

I need the area of the below shape

sukhopar [10]

The surface area of the triangular prism is 16cm

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3 years ago

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The menendez family paid $45 for a meal at a restaurant.They left a tip that was 20% of the cost of the meal.How much was the ti

horsena [70]

Answer:

Step-by-step explanation:

20% is equal to 1/5. Multiply $45 by 1/5 to get $9

8 0

3 years ago

What is the answer to this (in attached photo)

Katena32 [7]

Answer:

Step-by-step explanation:

Here is a list of each of the numbers until their factors add up to the same multiple:

Brendan - 4, 8, 12

Miguel - 6, 12

Jeron - 3, 6, 9, 12

- - - - -

Brendan would complete a lap about 3 times around 12 minutes.

Miguel would complete a lap about 2 times around 12 minutes.

Jeron would complete a lap about 4 times around 12 minutes.

IM CONCLUSION -

The answer is 12 minutes

5 0

3 years ago

A 1/17th scale model of a new hybrid car is tested in a wind tunnel at the same Reynolds number as that of the full-scale protot

Olegator [25]

Answer:

The ratio of the drag coefficients $\dfrac{F_m}{F_p}$ is approximately 0.0002

Step-by-step explanation:

The given Reynolds number of the model = The Reynolds number of the prototype

The drag coefficient of the model, $c_{m}$ = The drag coefficient of the prototype, $c_{p}$

The medium of the test for the model, $\rho_m$ = The medium of the test for the prototype, $\rho_p$

The drag force is given as follows;

$F_D = C_D \times A \times \dfrac{\rho \cdot V^2}{2}$

We have;

$L_p = \dfrac{\rho _p}{\rho _m} \times \left(\dfrac{V_p}{V_m} \right)^2 \times \left(\dfrac{c_p}{c_m} \right)^2 \times L_m$

Therefore;

$\dfrac{L_p}{L_m} = \dfrac{\rho _p}{\rho _m} \times \left(\dfrac{V_p}{V_m} \right)^2 \times \left(\dfrac{c_p}{c_m} \right)^2$

$\dfrac{L_p}{L_m} =\dfrac{17}{1}$

$\therefore \dfrac{L_p}{L_m} = \dfrac{17}{1} =\dfrac{\rho _p}{\rho _p} \times \left(\dfrac{V_p}{V_m} \right)^2 \times \left(\dfrac{c_p}{c_p} \right)^2 = \left(\dfrac{V_p}{V_m} \right)^2$

$\dfrac{17}{1} = \left(\dfrac{V_p}{V_m} \right)^2$

$\dfrac{F_p}{F_m} = \dfrac{c_p \times A_p \times \dfrac{\rho_p \cdot V_p^2}{2}}{c_m \times A_m \times \dfrac{\rho_m \cdot V_m^2}{2}} = \dfrac{A_p}{A_m} \times \dfrac{V_p^2}{V_m^2}$

$\dfrac{A_m}{A_p} = \left( \dfrac{1}{17} \right)^2$

$\dfrac{F_p}{F_m} = \dfrac{A_p}{A_m} \times \dfrac{V_p^2}{V_m^2}= \left (\dfrac{17}{1} \right)^2 \times \left( \left\dfrac{17}{1} \right) = 17^3$

$\dfrac{F_m}{F_p} = \left( \left\dfrac{1}{17} \right)^3$ = (1/17)^3 ≈ 0.0002

The ratio of the drag coefficients $\dfrac{F_m}{F_p}$ ≈ 0.0002.

5 0

3 years ago