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Anni [7]
3 years ago
6

PLEASE HELP ASAP!!

Mathematics
1 answer:
sesenic [268]3 years ago
8 0

f(x) =  {(x +  \frac{b}{2})}^{2}  +  \frac{24 - b}{2}

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Fin the values of x and y, help me
astraxan [27]

Answer:

x = 25 , y = 19

Step-by-step explanation:

Since the triangle has 3 congruent sides then it is equilateral

The 3 angles are also congruent, with each angle = 60° , then

2x + 10 = 60 ( subtract 10 from both sides )

2x = 50 ( divide both sides by 2 )

x = 25

and

3y + 3 = 60 ( subtract 3 from both sides )

3y = 57 ( divide both sides by 3 )

y = 19

3 0
3 years ago
In which quadrant is the number 6 – 8i located on the complex plane?
Bogdan [553]

Answer:

4th quadrant

Step-by-step explanation:

8 0
3 years ago
What is the unit price for trail mix if a 8 ounce bag cost $2.80.
german

Answer:

Step-by-step explanation:

2.80/8=.35 per ounce

6 0
2 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
John has $220 in his account and adds $25 to it each month Cheryl has $100 in her account and adds $35 each to her account. How
sammy [17]

Answer: In 12 months they will have the same amount, which will be $520.

Step-by-step explanation:

220 + 25m = 100 + 35m

m = month

Subtract 100 from each side

120 + 25m = 35m

Subtract 25m from each side

120 = 10m

Divide each side by 10

12 = m

4 0
3 years ago
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