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ehidna [41]
3 years ago
15

Suppose you sequentially add 10 drops of water to a graduated cylinder and read the volume add 10 more drops read and so on. how

many total drops will have been added after you do this four times
Mathematics
1 answer:
aleksandr82 [10.1K]3 years ago
3 0
It depends on what is being referred to by "this."

The description of the activity is that 10 drops are added twice and two readings are taken. If you do *that* 4 times, then a total of 20*4 = 80 drops will have been added.

If the activity referred to by "this" is the pair of acts {add 10 drops, read the volume}, then four repetitions will have added 10*4 = 40 drops.

Probably the expected interpretation is that "this" is {add 10 drops, read}, so 40 drops is likely the expected answer.
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Divide 55 by 5. get 11 times that by 4. you're welcome

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Imogene invested $8,000 in a bank account that pays 8 percent simple interest at the end of each year. Her friend invested the s
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The right system of equations to describe the situation would be on the form:

x1 = 8000 + y1*t

and

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Now for the value of amount earned, y1 and y2:

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3 years ago
Suppose an airplane climbs 15 feet for every 40 feet it moves forward. What is the slope of this airplanes ascent?
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Answer:

3/8

Step-by-step explanation:

Slope is the same as rise over run, or y/x. In this case, rise over run is 15 over 40, or 15/40. Simplified, this is 3/8.

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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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Answer:

Jessica hit the golf ball 20.50 yards farther than Kayla.

Step-by-step explanation:

150.75 - 130.25 = 20.50 yards

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