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lutik1710 [3]
2 years ago
7

Convert the decimal below to a fraction in simplest form . 0.41

Mathematics
1 answer:
Rufina [12.5K]2 years ago
7 0

Answer:

41/100

Step-by-step explanation:

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The value of y varies jointly with x and z. If y = 7 when z = 196 and x = 2, find the value of y when x = 3 and z = 336. I will
eimsori [14]

Answer:

18

Step-by-step explanation:

Given that:

y∞ xz

y=kxz. Where k is constant

When z=196 and x= 2 then y= 7

7=(196)(2)k

7=392k

k=1/56

There fore y=(1/56)xz

When x=3 and z =336

y=(1/56)xz

y=(1/56)(336)(3)

y=18

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At the city museum, child admission is $5.30 and adult admission is $9.20 . On Monday, 166 tickets were sold for a total sales o
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C+a=166
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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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