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Ilia_Sergeevich [38]
2 years ago
5

The centre of the circumstancribing circle of a triangle can be found by using the

Engineering
1 answer:
liraira [26]2 years ago
7 0

Answer:

perpendicular bisectors of each side of the triangle I believe

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Which of the following is not a function of the cooling system
Pie

Answer:

hola soy dora

Explanation:

5 0
2 years ago
John has just graduated from State University. He owes $35,000 in college loans, but he does not have a job yet. The college loa
IRISSAK [1]

Answer:

The correct response is "821.88". A further explanation is given below.

Explanation:

According to the question,

The largest amount unresolved after five years would have been:

= 35000\times (\frac{F}{P}, 4 \ percent,5 )

= 35000\times 1.216 7

= 42584.50

Now,

time (t) will be:

= 5\times 12

= 60 \ monthly \ payments

So, monthly payment will be:

= 42582.85\times (\frac{A}{P}, 0.5 \ percent,60 )

= 42584.50\times 0.0193

= 821.88

6 0
3 years ago
A closed, rigid tank is lled with a gas modeled as an ideal gas, initially at 27°C and a gage pressure of 300 kPa. The gas is he
sergejj [24]

Answer:

T₂ =93.77  °C

Explanation:

Initial temperature ,T₁ =27°C= 273 +27 = 300 K

We know that

Absolute pressure = Gauge pressure + Atmospheric pressure

Initial pressure ,P₁ = 300+1=301 kPa

Final pressure  ,P₂= 367+1 = 368  kPa

Lets take  temperature=T₂

We know that ,If the volume of the gas is constant ,then we can say that

\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}

{T_2}=T_1\times \dfrac{P_2}{P_1}

Now by putting the values in the above equation we get

{T_2}=300\times \dfrac{368}{301}\ K

The temperature in  °C

T₂ = 366.77 - 273  °C

T₂ =93.77  °C

8 0
3 years ago
A well is located in a 20.1-m thick confined aquifer with a conductivity of 14.9 m/day and a storativity of 0.0051. If the well
ahrayia [7]

Answer:

S = 5.7209 M

Explanation:

Given data:

B = 20.1 m

conductivity ( K ) = 14.9 m/day

Storativity  ( s ) = 0.0051

1 gpm = 5.451 m^3/day

calculate the Transmissibility ( T ) = K * B

                                                       = 14.9 * 20.1 = 299.5  m^2/day

Note :

t = 1

U = ( r^2* S ) / (4*T*<em> t </em>)

  = ( 7^2 * 0.0051 ) / ( 4 * 299.5 * 1 ) = 2.0859 * 10^-4

Applying the thesis method

W(u) = -0.5772 - In(U)

       = 7.9

next we calculate the pumping rate from well ( Q ) in m^3/day

= 500 * 5.451 m^3 /day

= 2725.5 m^3 /day

Finally calculate the drawdown at a distance of 7.0 m form the well after 1 day of pumping

S = \frac{Q}{4\pi T} * W (u)

 where : Q = 2725.5

               T = 299.5

               W(u)  = 7.9

substitute the given values into equation above

S = 5.7209 M

4 0
3 years ago
Cuál de las siguientes es la mejor manera de practicar sus habilidades de tecnología de secundaria?
Mkey [24]
Huh? Do you know English?
8 0
3 years ago
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