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Dmitriy789 [7]
3 years ago
13

Find x. Please help !!

Mathematics
1 answer:
Korvikt [17]3 years ago
3 0
I am pretty sure it’s C but I’m not very good at these kind of problems...
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12³.2².4⁴ / 4³.2.12²
Nuetrik [128]
Remember
\frac{xyz}{abc}=( \frac{x}{a} )( \frac{y}{b} )( \frac{z}{c} )
and
\frac{x^m}{x^n}=x^{m-n}

split them to have same bases
\frac{12^32^24^4}{4^32^112^2}=
( \frac{12^3}{12^2} )( \frac{4^4}{4^3} )( \frac{2^2}{2^1} )=
(12^1)(4^1)(2^1)=
(12)(4)(2)=
96
8 0
3 years ago
Read 2 more answers
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
What is the range of the data in this stem-and-leaf plot?
sergij07 [2.7K]
The range is the difference between the lowest and the highest number.

the answer is 30.5
3 0
4 years ago
Read 2 more answers
(35 points) Hey please help quickly. :(
UkoKoshka [18]

Answer:here ya go boys

Step-by-step explanation:

6 0
3 years ago
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Suppose that the following group of values has been entered into the TVM
Mila [183]

Answer:

The answer on A P E X is bal(132)

5 0
3 years ago
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