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Kaylis [27]
2 years ago
12

Tom has been a customer of the same bank foe two years and his income can change frommonth to month after paying bills. Tom plac

es 250.00 a month in a two year CD. He notices that last year twice he withdrew 250.00 from his CD early to pay monthly bills. His bank charges a penalty for a early CD withdraw. His bank requires a minimum of 150.00 for regular savings accounts 750.00 for money markets, and 1,000 for CDs what would be the best strategy
Mathematics
1 answer:
atroni [7]2 years ago
8 0

Answer:

i thick Tom money will be less, because the bank 150.00 from his saving acount.

250-150.00=100

his money will be decreasing

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3

Step-by-step explanation:

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13. AC=6 

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What is the equation for the kind of reflection?
andre [41]

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A vertical reflection is given by the equation y=−f(x) y = − f ( x ) and results in the curve being “reflected” across the x-axis. A horizontal reflection is given by the equation y=f(−x) y = f ( − x ) and results in the curve being “reflected” across the y-axis.

4 0
2 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
Helpppppppppppppppppp
makvit [3.9K]

Answer:

c. 25

Step-by-step explanation:

100/12 = 8.3

8.3 × 3 = 24.9

rounded up is 25

6 0
3 years ago
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