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Anni [7]
3 years ago
6

Which term is the perfect square of 3x4?

Mathematics
2 answers:
mr Goodwill [35]3 years ago
7 0

Answer:

The correct answer is C) 9x^8

Step-by-step explanation:

When multiplying, you add the exponents, so therefore the correct answer woulf be 9x^8.

Which term is the perfect square of 3x4?

1) multiply

<em>(3x^4)*(3x^4)</em>

2) simpilify

<em>3*3= 9</em>

<em>4+4= 8</em>

So you answer would be 9x^8

Alexus [3.1K]3 years ago
6 0
9*16
                        vjgfckhgckfgxkgfxgfxkhfx                
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Assume y varies directly as x. if y=8 when x=42; find y when x=126
Soloha48 [4]
Varies directly basically means a multiple of

y = kx

where "k" is the constant of variation. You must solve for this first using the given information.

8 = 42k
8/42 = k
0.19= k

Equation: y = 0.19x
when x = 126
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6 0
3 years ago
A tank contains 5,000 L of brine with 13 kg of dissolved salt. Pure water enters the tank at a rate of 50 L/min. The solution is
tresset_1 [31]

Answer:

a) x(t) = 13*e^(^-^\frac{t}{100}^)

b) 10.643 kg

Step-by-step explanation:

Solution:-

- We will first denote the amount of salt in the solution as x ( t ) at any time t.

- We are given that the Pure water enters the tank ( contains zero salt ).

- The volumetric rate of flow in and out of tank is V(flow) = 50 L / min  

- The rate of change of salt in the tank at time ( t ) can be expressed as a ODE considering the ( inflow ) and ( outflow ) of salt from the tank.

- The ODE is mathematically expressed as:

                            \frac{dx}{dt} = ( salt flow in ) - ( salt flow out )

- Since the fresh water ( with zero salt ) flows in then ( salt flow in ) = 0

- The concentration of salt within the tank changes with time ( t ). The amount of salt in the tank at time ( t ) is denoted by x ( t ).

- The volume of water in the tank remains constant ( steady state conditions ). I.e 10 L volume leaves and 10 L is added at every second; hence, the total volume of solution in tank remains 5,000 L.

- So any time ( t ) the concentration of salt in the 5,000 L is:

                             conc = \frac{x(t)}{1000}\frac{kg}{L}

- The amount of salt leaving the tank per unit time can be determined from:

                         salt flow-out = conc * V( flow-out )  

                         salt flow-out = \frac{x(t)}{5000}\frac{kg}{L}*\frac{50 L}{min}\\

                         salt flow-out = \frac{x(t)}{100}\frac{kg}{min}

- The ODE becomes:

                               \frac{dx}{dt} =  0 - \frac{x}{100}

- Separate the variables and integrate both sides:

                       \int {\frac{1}{x} } \, dx = -\int\limits^t_0 {\frac{1}{100} } \, dt  + c\\\\Ln( x ) = -\frac{t}{100} + c\\\\x = C*e^(^-^\frac{t}{100}^)

- We were given the initial conditions for the amount of salt in tank at time t = 0 as x ( 0 ) = 13 kg. Use the initial conditions to evaluate the constant of integration:

                              13 = C*e^0 = C

- The solution to the ODE becomes:

                           x(t) = 13*e^(^-^\frac{t}{100}^)

- We will use the derived solution of the ODE to determine the amount amount of salt in the tank after t = 20 mins:

                           x(20) = 13*e^(^-^\frac{20}{100}^)\\\\x(20) = 13*e^(^-^\frac{1}{5}^)\\\\x(20) = 10.643 kg

- The amount of salt left in the tank after t = 20 mins is x = 10.643 kg

                           

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