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kolbaska11 [484]
3 years ago
11

What would you have to do to change 10 cubic feet into cubic inches? A. Multiply by 46,656 B. Divide by 1,728 C. Multiply by 1,7

28 D. Divide by 46,656
Mathematics
2 answers:
Vinil7 [7]3 years ago
7 0
C. Multiply by 1,728 is the answer
Semmy [17]3 years ago
5 0
1 foot = 12 inches

(1 foot)³ = (12 inches)³

12³ = 1,728 in³

To change 10 cubic feet into cubic inches

10 ft³ * 1728 in³/ft³ = 17,280 in³  Choice C. Multiply by 1,728
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Hi could someone tell me how to solve this, you don’t even have to tell me the answer I’m just confused on this math and need he
BabaBlast [244]

Answer:

132 ft

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8 0
3 years ago
Read 2 more answers
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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sertanlavr [38]

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Bingel [31]
I don’t think f would have any value in this equation and f would represent 1 zero??
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