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4 years ago

A rectangle has a length 2 ft less than three times it's width. What is the area of the rectangle when it's width is 2ft?

Mathematics

1 answer:

Andreyy894 years ago

4 0

Answer:

The area is 4 ft 2 inches

Step-by-step explanation:

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What percentage is 26 out of 130?

OLga [1]

Answer: 20%

<em>I hope this helps, and Happy Holidays! :)</em>

7 0

3 years ago

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If g(x)=3x-11, what is x is g(x)=289

Liula [17]

Answer:x=100

Step-by-step explanation:

Since you know that g(x)=3x-11 all you have to do is substitute that into g(x)=289. So you get 3x-11=289. To get x you have to add 11 on both sides.

3x-11=289

+11 +11

----------------

3x = 300

Divide 3 on both sides to get x by itself

3x =300

---- -------

3 3

x=100

8 0

3 years ago

What is the length, in centimeters (cm), of y?

Andre45 [30]

Answer:

should be 6

Step-by-step explanation:

Pythagorean Theorum was used.

6 0

3 years ago

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Solve for x<br> Answer must be simplified<br> x - 6 ≤ 3

Usimov [2.4K]

Answer:

x ≤ 9

Step-by-step explanation:

x - 6 ≤ 3

+ 6 + 6

X ≤ 9

6 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