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Ainat [17]
3 years ago
10

As a team leader, Lisa shows compassion to her subordinates and respects them. She gives credit to her followers for their achie

vements. She is generally positive about the operations of her team and spreads warmth and energy among her team members. This indicates that Lisa is which type of leader?
Mathematics
1 answer:
svet-max [94.6K]3 years ago
3 0

Answer: This indicates that Lisa is a Person-Oriented leader, which keeps her subordinates motivated and can influence their beliefs in  a positive way, so they become more productive and involved in the process. She works with people's self-esteem so they become <u>happier and more confident</u> in their jobs, keeping up the perfomance. This kind of leader uses charisma in order to keep their employees morale up.

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If f(1) = 2 and f(n) = f(n − 1)2 + 3 then find the value of f(3).
blondinia [14]

Answer:

52

Step-by-step explanation:

f(n) is purely based on the previous value of f(n), or f(n-1), so we can start with f(1) and work our way up. We know f(1) = 2, so to find f(2), we plug f(1) into

f(n-1)²+3 to get

f(1)²+3 = 2²+3 = 4+3=7

Thus, f(2) =7. Similarly,

f(3) = f(3-1)²+3 = f(2)² + 3 -= 7² + 3= 52

3 0
2 years ago
Five rational numbers between -1 and -2
Vika [28.1K]

Answer:

-1/2, -1/3,-1/4,-1/5

6 0
3 years ago
The use of mathematical methods to study the spread of contagious diseases goes back at least to some work by Daniel Bernoulli i
harina [27]

Answer:

a

   y(t) = y_o e^{\beta t}

b

      x(t) =  x_o e^{\frac{-\alpha y_o }{\beta }[e^{-\beta t} - 1] }

c

      \lim_{t \to \infty} x(t) = x_oe^{\frac{-\alpha y_o}{\beta } }

Step-by-step explanation:

From the question we are told that

    \frac{dy}{y} =  -\beta dt

Now integrating both sides

     ln y  =  \beta t + c

Now taking the exponent of both sides

       y(t) =  e^{\beta t + c}

=>     y(t) =  e^{\beta t} e^c

Let  e^c =  C

So

      y(t) = C e^{\beta t}

Now  from the question we are told that

      y(0) =  y_o

Hence

        y(0) = y_o  = Ce^{\beta * 0}

=>     y_o = C

So

        y(t) = y_o e^{\beta t}

From the question we are told that

      \frac{dx}{dt}  = -\alpha xy

substituting for y

      \frac{dx}{dt}  = - \alpha x(y_o e^{-\beta t })

=>   \frac{dx}{x}  = -\alpha y_oe^{-\beta t} dt

Now integrating both sides

         lnx = \alpha \frac{y_o}{\beta } e^{-\beta t} + c

Now taking the exponent of both sides

        x(t) = e^{\alpha \frac{y_o}{\beta } e^{-\beta t} + c}

=>     x(t) = e^{\alpha \frac{y_o}{\beta } e^{-\beta t} } e^c

Let  e^c  =  A

=>  x(t) =K e^{\alpha \frac{y_o}{\beta } e^{-\beta t} }

Now  from the question we are told that

      x(0) =  x_o

So  

      x(0)=x_o =K e^{\alpha \frac{y_o}{\beta } e^{-\beta * 0} }

=>    x_o = K e^{\frac {\alpha y_o  }{\beta } }

divide both side  by    (K * x_o)

=>    K = x_o e^{\frac {\alpha y_o  }{\beta } }

So

    x(t) =x_o e^{\frac {-\alpha y_o  }{\beta } } *  e^{\alpha \frac{y_o}{\beta } e^{-\beta t} }

=>   x(t)= x_o e^{\frac{-\alpha * y_o }{\beta} + \frac{\alpha y_o}{\beta } e^{-\beta t} }

=>    x(t) =  x_o e^{\frac{\alpha y_o }{\beta }[e^{-\beta t} - 1] }

Generally as  t tends to infinity ,  e^{- \beta t} tends to zero  

so

    \lim_{t \to \infty} x(t) = x_oe^{\frac{-\alpha y_o}{\beta } }

5 0
3 years ago
Please answer this problem!
harina [27]

Answer:

a. volume= 784 is the answer

8 0
2 years ago
Alana and Ramiro are writing an article for the school newspaper about whether their classmates prefer to buy their lunch or bri
MrRissso [65]

Answer:Ramiro’s method is

Step-by-step explanation:

Did the assignment

3 0
3 years ago
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