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sesenic [268]
3 years ago
13

9[n+2(n-2)]=45how do u answer this question it has to many numbers.​

Mathematics
1 answer:
liq [111]3 years ago
3 0

Answer:

n = 3

Step-by-step explanation:

9{n+2(n-2)} = 45

Start with the numbers on the inner most pair of parentheses

Multiply (n-2) by 2

9[n+2n-4] = 45

Add what is inside the parentheses

9[3n-4] = 45

Multiply

27n-36 = 45

Add 36 to both sides

27n = 81

Divide both sides by 27

n = 3

Hope this helps!

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WORTH 10 points.
Tanya [424]
The answer is B because I did the math
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3 years ago
Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another br
Lina20 [59]

Answer:

The differential equation for the amount of salt A(t) in the tank at a time  t > 0 is \frac{dA}{dt}=12 - \frac{2A(t)}{500+t}.

Step-by-step explanation:

We are given that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another brine solution is pumped into the tank at a rate of 3 gal/min, and when the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

The concentration of the solution entering is 4 lb/gal.

Firstly, as we know that the rate of change in the amount of salt with respect to time is given by;

\frac{dA}{dt}= \text{R}_i_n - \text{R}_o_u_t

where, \text{R}_i_n = concentration of salt in the inflow \times input rate of brine solution

and \text{R}_o_u_t = concentration of salt in the outflow \times outflow rate of brine solution

So, \text{R}_i_n = 4 lb/gal \times 3 gal/min = 12 lb/gal

Now, the rate of accumulation = Rate of input of solution - Rate of output of solution

                                                = 3 gal/min - 2 gal/min

                                                = 1 gal/min.

It is stated that a large mixing tank initially holds 500 gallons of water, so after t minutes it will hold (500 + t) gallons in the tank.

So, \text{R}_o_u_t = concentration of salt in the outflow \times outflow rate of brine solution

             = \frac{A(t)}{500+t} \text{ lb/gal } \times 2 \text{ gal/min} = \frac{2A(t)}{500+t} \text{ lb/min }

Now, the differential equation for the amount of salt A(t) in the tank at a time  t > 0 is given by;

= \frac{dA}{dt}=12\text{ lb/min } - \frac{2A(t)}{500+t} \text{ lb/min }

or \frac{dA}{dt}=12 - \frac{2A(t)}{500+t}.

4 0
3 years ago
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Alika [10]

9514 1404 393

Answer:

  y +4 = -5/14(x -7)

Step-by-step explanation:

The slope can be found using the slope formula:

  m = (y2 -y1)/(x2 -x1)

  m = (1 -(-4))/(-7-7) = 5/-14 = -5/14

The point-slope form is ...

  y -k = m(x -h) . . . . . line with slope m through point (h, k)

Filling in the given values, we have ...

  y +4 = -5/14(x -7)

7 0
3 years ago
The sanchez family goes out for dinner, and the price of the meals is $60. The sales tax on the Meals is 7 percent, and they als
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Answer:

73.3

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3 years ago
Helppppp giving brainly too ​
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Alright. (1/4^2)+2((1/4)(2/3)-3(2/3^2)=-.9375 is the correct answer
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3 years ago
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