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3 years ago

PLEASE EXPLAIN!!! NEED HELP!!

1 answer:

3 0

We will say 5 units=3 dollars
1/2 of 5 units=3 dollars (just so I don't get confused)
to find 1/3
times both sides by 2/3 to get
1/3 of 5 units=2 dollars
then times 2 both sides
1/3 of 10 units=4 dollars

so the value of 1/3 of 10 is 4

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Please help!<3 tysm if u helped

Vilka [71]

Answer:

x = 0 1 -1 -2

y = -3 -4 -2 -1

Step-by-step explanation:

equation : 3y+9x = -9

when x = 1

3y+3x = -9

or, 3y + 3×1= -9

or, 3y+3=-9

or, 3y = -12

so, y = -4

Again

Let's suppose x as -1

3y+3x= -9

or, 3y+ 3×-1 = -9

or, 3y-3 = -9

or, 3y = -6

y = -2

Let's suppose x as -2 now,

3y+3x = -9

3y+3×-2 = -9

or, 3y-6 = -9

or, 3y = -3

so, y = -1

5 0

2 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;

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

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

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

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

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