Streaklines are traced out by neutrally buoyant marker fluid injected into a flow field from a fixed point in space. A particle of the marker fluid that is at point ðx, yÞ at time t must have passed through the injection point ðx0,y0Þ at some earlier instant t =τ. The time history of a marker particle may be found by solving the pathline equations for the initial conditions that x=x0, y=y0 when t =τ. The present locations of particles on the streakline are obtained by setting τ equal to values in the range 0≤τ ≤t. Consider the flow field V! =axð1+btÞ^i+cy^j, where a=c=1 s−1 and b=0:2 s−1. Coordinates are measured in meters. Plot the streakline that passes through the initial point ðx0,y0Þ=ð1,1Þ, during the interval from t =0 to t =3 s. Compare with the streamline plotted through the same point at the instants t =0, 1, and 2 s.

the solution of the equation is 2.23 ⇒ answer b

step-by-step explanation:

* lets revise the meaning of exponential function

- the form of the exponential function is y = ab^x, where a ≠ 0, b > 0 ,  

  b ≠ 1, and x is any real number

- it has a constant base b

- it has a variable exponent x

- to solve an exponential equation, take the log or ln of both sides,  

  and solve for the variable

* lets solve the problem

∵

- the base is 7 and the exponent is x

- insert ㏑ in both sides

∴

- use the rule

∴ x ㏑(7) = ㏑(77)

- to find x divide both sides by ㏑(7)

∴

* the solution of the equation is 2.23 to the nearest 2 decimal places

the polynomial is a perfect square trinomial of the form a^2-2ab+b^2:

option a. true

step-by-step explanation:

9x^2-12x+4

comparing with the form a^2-2ab+b^2

1.) a^2=9x^2

solving for a: square root both sides of the equation:

sqrt(a^2)=sqrt(9x^2)

a=sqrt(9) sqrt(x^2)

a=3x

2.) b^2=4

solving for b: square root both sides of the equation:

sqrt(b^2)=sqrt(4)

b=2

to be a perfet square trinomial 2ab must be equal to 12x. let's check:

2ab = 2 (3x) (2)

2ab = 12x

correct

the answer would be  

a.

3: 1

step-by-step explanation: