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2 years ago

Can someone explain how to calculate a unit rate

Mathematics

1 answer:

MArishka [77]2 years ago

5 0

Answer:

It's like 60 miles per (1) hour or 20 mL per (1) second. Independent value has to equal 1.

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Figure ABCD is a rectangle Find the value of x

romanna [79]

Step-by-step explanation:

Since this is a rectangle these two equations for the length have to equal each other.

2x+4=7x-1

We now need to get all of the x's on one side and constant numbers on the other side.

add 1 to both sides and subtract 2x from both sides.

5=5x

divide 5 from both sides.

x=1

6 0

2 years ago

According to the table, what was the hiker’s total displacement?

shtirl [24]

Answer:

A: 0

Step-by-step explanation:

Important to remember - total displacement is the direct distance between the start and finish points, and isn't always equal to the total distance travelled.

In the case of the table above, the hiker travels the same distance in opposite directions, which actually means they cancel each other out! Traveling 4 km to the west, then 4 km to east means you end up in the same spot, right? So your total displacement is 0! (or 4 - 4, which equals 0). Same happens with the north-south hikes - the hiker ends up in the same spot, as the displacement is equal to 6 - 6 = 0.

So, the total displacement of the hiker is 0. (Even though the hiker travelled 20 km!)

7 0

2 years ago

Write an equivalent expression for -6(a+8)

stepan [7]

-6a-48 that would be it because you are only distributing the -6

8 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;

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

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

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

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

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