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jarptica [38.1K]
3 years ago
14

What should -7/56 BE DIVIDE TO GET -1/8​

Engineering
2 answers:
Luba_88 [7]3 years ago
8 0

Answer:

i dont remember how to do this im so sorry but i think its -14.5

Explanation:

aivan3 [116]3 years ago
5 0

Answer:

7/7

Explanation:

7/56 ÷1

7/56÷ 7/7

7/56 ×7/7

1/8

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Five hundred gallons of 89-octane gasoline is obtained by mixing 87-octane gasoline with 92-octane gasoline. (a) Write a system
miskamm [114]

Explanation:

a) The total volume equals the sum of the volumes.

500 = x + y

The total octane amount equals the sum of the octane amounts.

89(500) = 87x + 92y

44500 = 87x + 92y

b) desmos.com/calculator/ekegkzllqx

As x increases, y decreases.

c) Use substitution or elimination to solve the system of equations.

44500 = 87x + 92(500−x)

44500 = 87x + 46000 − 92x

5x = 1500

x = 300

y = 200

The required volumes are 300 gallons of 87 gasoline and 200 gallons of 92 gasoline.

6 0
3 years ago
What is this spray pattern defect most likely caused by:
DiKsa [7]

Answer:

fluid nozzle that is too large

6 0
2 years ago
A heat pump operates on a Carnot heat pump cycle with a COP of 12.5. It keeps a space at 24°C by consuming 2.15 kW of power. Det
Vinil7 [7]

Answer:

a) T_{L} = 273.378\,K\,(0.228\,^{\textdegree}C), b) \dot Q_{H} = 26.875\,kW

Explanation:

a) The Coefficient of Performance of the Carnot Heat Pump is:

COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}

After some algebraic handling, the temperature of the cold reservoir is determined:

T_{H}-T_{L} = \frac{T_{H}}{COP_{HP}}

T_{L} = T_{H}\cdot \left(1-\frac{1}{COP_{HP}}  \right)

T_{L} = (297.15\,K)\cdot \left(1-\frac{1}{12.5}\right)

T_{L} = 273.378\,K\,(0.228\,^{\textdegree}C)

b) The heating load provided by the heat pump is:

\dot Q_{H} = COP_{HP}\cdot \dot W

\dot Q_{H} = (12.5)\cdot (2.15\,kW)

\dot Q_{H} = 26.875\,kW

4 0
3 years ago
Which rigid motion maps the solid-line figure onto the dotted-line figure?
Agata [3.3K]
I would love to answer but unfortunately there is no picture.
5 0
3 years ago
In a certain chemical plant, a closed tank contains ethyl alcohol to a depth of 71 ft. Air at a pressure of 17 psi fills the gap
Yuliya22 [10]

Answer:

the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi

Explanation:

Given that;

depth 1 = 71 ft

depth 2 = 10 ft

pressure p = 17 psi = 2448 lb/ft²

depth h = 71 ft - 10 ft = 61 ft

we know that;

p = P_air + yh

where y is the specific weight of ethyl alcohol ( 49.3 lb/ft³ )

so we substitute;

p = 2448 + ( 49.3 × 61 )

= 2448 + 3007.3

= 5455.3 lb/ft³

= 37.88 psi

Therefore, the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi

5 0
3 years ago
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