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DerKrebs [107]
3 years ago
8

What is the solution to the system of equations?2x-y=7y=2x+3​

Engineering
1 answer:
AleksAgata [21]3 years ago
6 0

Answer:

No solution.

Explanation:

Solve with substitution or elimination.

If you use substitution, plug the expression for y in the second equation into the first equation.

2x − (2x + 3) = 7

2x − 2x − 3 = 7

-3 = 7

No solution.

If you use elimination, add the equations together to eliminate y.

2x − y + y = 7 + 2x + 3

2x = 10 + 2x

0 = 10

No solution.

There is no solution.

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What are the 3 dimensions that used in isometric sketches?
noname [10]

Answer:

The three dimensions shown in an isometric drawing are the height, H, the length, L, and the depth, D

Explanation:

An isometric drawing of an object in presents a pictorial projection of the object in which the three dimension, views of the object's height, length, and depth, are combined in one view such that the dimensions of the isometric projection drawing are accurate and can be measured (by proportion of scale) to draw the different views of the object or by scaling, for actual construction of the object.

5 0
3 years ago
Does anybody know what plane this is? i saw it the other day doing a low pass through my community
Arlecino [84]

Answer:

Airbus A340-313

Explanation:

it is what it is

3 0
2 years ago
A factory hires a consultant to recommend ways to improve its productivity. The consultant notices that the production floor is
Romashka [77]

Answer:

B

Explanation:

Keep only what is used. Remove all unnecessary items. Red tag these items for review.

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2 years ago
Oil with a density of 850 kg/m3 and kinematic viscosity of 0.00062 m2 /s is being discharged by a 5 mm diameter, 40 m long horiz
ra1l [238]

Answer:

Flow rate is 1.82\times 10^{-8} m^{3}/s

Explanation:

Given information

Density of oil, \rho_{oil}= 850 Kg/m^{3}

kinematic viscosity, v= 0.00062 m^{2} /s

Diameter of pipe, D= 5 mm= 0.005 m

Length of pipe, L=40 m

Height of liquid, h= 3 m

Volume flow rate for horizontal pipe will be given by

\bar v=\frac {\triangle P\pi D^{4}}{128\mu L} where \mu is dynamic viscosity and \triangle P is pressure drop

At the bottom of the tank, pressure is given by

P_{bottom}=\rho_{oil} gh=850 Kg/m^{3}\times 9.81 m/s^{2}\times 3 m= 25015.5 N/m^{2}

Since at the top pressure is zero, therefore change in pressure a difference between the pressure at the bottom and the top. It implies that change in pressure is still 25015.5 N/m^{2}

Dynamic viscosity, \mu=\rho_{oil}v= 850 Kg/m^{3}\times 0.00062 m^{2}/s=0.527 Kg/m.s

Now the volume flow rate will be

\bar v=\frac {25015.5 N/m^{2}\times \pi \times 0.005^{4}}{128\times 0.527 Kg/m.s \times 40}=1.82037\times 10^{-8} m^{3}/s\approx 1.82\times 10^{-8} m^{3}/s

Proof of flow being laminar

The velocity of flow is given by

V_{flow}=\frac {\bar v}{A}=\frac {1.82\times 10^{-8} m^{3}/s}{0.25\times \pi\times 0.005^{2}}=0.000927104  m/s

Reynolds number, Re=\frac {\rho_{oil} v_{flow} D}{\mu}=\frac {850 Kg/m^{3}\times 0.000927104 m/s\times 0.005}{0.527 kg/m.s}=0.007476648

Since the Reynolds number is less than 2300, the flow is laminar and the assumption is correct.

5 0
3 years ago
A shaft machined of AISI 1022 steel having σuts = 70 ksi and σy = 52 ksi is loaded cyclically. The loading is characterized by t
Maru [420]

Answer:

2.075 in

Explanation:

Please kindly check attachment for the detailed and step by step solution of the given problem.

3 0
2 years ago
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