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3 years ago

Which sequence of transformations can change the size of a polygon? Select all that apply

Mathematics

2 answers:

skelet666 [1.2K]3 years ago

5 0

Answer: C

Step-by-step explanation:

Kruka [31]3 years ago

4 0

C. Dilation and translation

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Which BEST describes a ray?

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Answer:

Endpoint at one end and an arrow at the other end.

Step-by-step explanation:

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3 years ago

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How does tripling the side lengths of a right triangle affect its area?)

kap26 [50]

The area of the triangle will then be 9 times as great.

The formula of the area of a triangle is 1/2bh, so if the base has been tripled and the height has been tripled, you're essentially multiplying 3 to the original equation twice (3 x 3 = 9).

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3 years ago

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Complete the missing value in the solution to the equation. <br><br> y = 3x -7<br><br> (1, ?,)

Luda [366]

Answer: -4

Step-by-step explanation:

So, for y=3x-7, substitute the x with 1 (from (1,?) ) which gives you y=3(1)-7. Lastly, when you solve the equation, it would give you y=-4.

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3 years ago

Simplify by combining like terms: -4x - 2y - 7x - 8y

Sav [38]

Step-by-step explanation:

Solving like terms

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4 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