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djverab [1.8K]
3 years ago
12

2−2n=3n+17 what IS THE answer help

Mathematics
2 answers:
Stolb23 [73]3 years ago
6 0
First, let's identify the like terms. 
2 and 17
2n and 3n

Next, you would need to combine them. Ex: Add 2n to both sides, and subtract 17 from both sides. 
2 - 2n = 3n + 17
2 = 5n + 17
-15 = 5n

Now, all you would need to do is isolate the n. To do this, you would divide both sides by 5. 
-15 = 5n
-3 = n

n = -3

The solution would be -3. 

I hope this helps!
BartSMP [9]3 years ago
6 0
★ LINEAR REDUCTION ★

Given equation :

2 - 2n = 3n + 17

- 3n - 2n = 17 - 2 = 15

- 1 ( 3n + 2n ) = 15

-1 ( 5n ) = 15

5n = - 15

Hence , n = -3

★✩★✩★✩★✩★✩★✩★✩★✩★✩★✩★
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5×12÷[-12÷{15+(4-13)}]​
Elina [12.6K]

Answer:

=5×12÷[-12÷{15+(-9)}]

=5×12÷[-12÷6]

=5×12÷(-2)

=-30

Step-by-step explanation:

8 0
3 years ago
Imagine you have a job where you earn $700 each week after taxes. Your monthly living expenses include $725 for rent, $125 for u
Softa [21]

The monthly income is $844.

Given:

  • Weekly earning after taxes is $700
  • Monthly rent is $725
  • Monthly expense for utilities is $125
  • Monthly expense for cable/internet is $120
  • Monthly expense for cellphone is $35
  • Monthly expense for bus fare is $48
  • Minimum monthly payment of credit card debt is $78
  • Monthly expense for groceries is $600
  • Monthly expense for dining out and entertainment is $225

To find: The monthly income

The monthly income refers to the amount of money left over each month after paying all expenses.

Evaluating, we have,

The total monthly expense = $(725 + 125 + 120 + 35 + 48 + 78 + 600 + 225)

That is, the total monthly expense is $1956.

It is given that weekly earning after taxes is $700. Since a month contains 4 weeks, we can say that monthly earning after taxes is $(700 \times 4), that is, $2800.

Then, monthly income is given by the difference between the monthly earnings and the monthly expenses. So, monthly income is $(2800-1956), that is, $844.

The monthly income is $844.

Learn more about income and expenses here:

brainly.com/question/18456246

7 0
3 years ago
What the answer to 5
Sonja [21]

Answer:

c. n³p²

Step-by-step explanation:

3 0
3 years ago
The bottom of a boat lles 7 feet below the surface of the water the boat stands 18 feet above the surface of the water.
ASHA 777 [7]

Answer:

18-(-7)

Step-by-step explanation:

7 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
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