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3 years ago

-1/3 = j/4 - 10/3 what is j

Mathematics

1 answer:

lesantik [10]3 years ago

6 0

Answer:

j=12

Step-by-step explanation:

Question: <u>j/4-10/3=-1/3</u>

1) Add 10/3 to both sides:

j/4=3

2) Multiply both sides by 4:

j=12

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Write an equation of the line that passes through 3,1 and 0,10

elena-14-01-66 [18.8K]

Answer:y = -3x + 10

Step-by-step explanation:

To find an equation of a line that passes through two points, we have to first find the slope between the two equation. We can do this by using the slope formula:

where (x₁, y₁) and (x₂, y₂) are the two points that we are finding the slope between.

Lets make (x₁, y₁) equal to (0, 10) and (x₂, y₂) equal to (3, 1). Now we plug them into the slope formula:

So the slope between the two points is -3.

From here, I would normally take one of the points given to us and plug in the point and slope into the point-slope form of a line and then simplify until we get it in slope-intercept form. But if you look carefully, the y-intercept is given to us as the point (0, 10). So we now know that the y-intercept of the line is 10. We can now take the y-intercept and the slope and plug it into the slope-intercept form of a line to get out equation:

y = mx + b

plug in -3 for m (the slope) and 10 for b (the y-intercept)

y = -3x + 10

So now we have our equation.

I hope you find my answer and explanation helpful. Happy studying. :)

3 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;

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

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

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

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

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