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2 years ago

The following excerpt comes from the international bottled water association:

Mathematics

1 answer:

lora16 [44]2 years ago

8 0

Using proportions, the estimate of the per capita consumption from 2011 was 29.25 gallons.

<h3>What is a proportion?</h3>

A proportion is a fraction of the total amount, the measures are related using a rule of three.

The amount in 2012 was an increase of 5.3% from 2011,

105.3% = 1.053 of x,

The consumption per capita was 30.8 gallons,

hence:

1.053x = 30.8

x = 30.8/1.053

x = 29.25 gallons.

More can be learned about proportions at brainly.com/question/24372153

#SPJ1

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3 years ago

The coordinates of the vertices of triangle ABC are A(-1,3), B(1,2) and C(-3,-1).

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3 years ago

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Xelga [282]

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

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3 years ago

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riadik2000 [5.3K]

The answer to #3 is 52.

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4 years ago