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2 years ago

387.5 divided by 2.5

Mathematics

2 answers:

tekilochka [14]2 years ago

7 0

Answer:

155

Step-by-step explanation:

ArbitrLikvidat [17]2 years ago

5 0

Answer:

155

Step-by-step explanation:

Use the calculator :D

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Plz help my teacher isn’t emailing my back

nalin [4]

Answer:

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Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

The sum of 4 consecutive integers is 26. What is the least of 4 integers?

Veseljchak [2.6K]

Answer:

Step-by-step explanation:

it is 5 because 5 + 6 + 7 + 8 =26

and 5 is the least

4 0

3 years ago

Read 2 more answers

What is the range of possible sides for side x? The other two sides are 3.2 and 5.5

Alecsey [184]

Answer:

2.3 < x < 8.7

Step-by-step explanation:

Given 2 sides of a triangle then the 3rd side x is in the range

difference of 2 sides < x < sum of 2 sides , that is

5.5 - 3.2 < x < 5.5 + 3.2

2.3 < x < 8.7

6 0

3 years ago

find five solutions of the equation y=20x. select integers for values for x starting with -2 and ending with 2

Nataly_w [17]

Answer:

-440, -4440, -44440, -444440, -4444440

Step-by-step explanation:

First, let's find values for x which start is a -2 and end with a 2.

-22

-222

-2222

-22222

-222222

Now we can find the y value by multiplying each term by 20.

-22 --> -440

-222 --> -4440

-2222 --> -44440

-22222 --> -444440

-222222 --> -4444440

7 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