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Margarita [4]
3 years ago
13

12/5 divided by 21/10

Mathematics
1 answer:
Juli2301 [7.4K]3 years ago
6 0
Exact form is 8/7 , as a decimal its 1.142 and mixed number form its 1 1/7
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Polygon ABCDE is the first in a pattern for a high school art project. The polygon is transformed so that the image of A’ is at
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Answer:

4.21

Step-by-step explanation:

Because 2.1 is also equivalent to  -4.2 which also is equivalent to D'.

7 0
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6n+3=2 use a table ....
kvasek [131]
You - 3 from 2 then do 6n=-1
6 0
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
2.3 repeating as a fraction​
Marat540 [252]
23/10 = 2 and 3/10




I hope this helped
7 0
3 years ago
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Select the two values of x that are roots of this equation.<br> x2 - 5x + 2 = 0
Margaret [11]

Answer:

C and D

{x}^{2}  - 5x + 2 = 0 \\ x =  \frac{5± \sqrt{17} }{2}

8 0
3 years ago
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