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Travka [436]
3 years ago
12

When dividing the polynomial p (1) = 4t^3 - 17t^2+ 14t - 3 by t - 3, the remainder can be determined by evaluating p(3). What is

the value of this
remainder?
A P(3) = -27
B. P(3) = -6
C. P(3) = 0
D. p(3) = 45
Mathematics
1 answer:
Brut [27]3 years ago
3 0

Answer:

The answer to your question is the letter B. P(3) = -6

Step-by-step explanation:

                                     4t²- 5t -1

                           t - 3   4t³ - 17t² + 14t - 3

                                   -4t³  +12t²

                                     0    -5t² +14t

                                           +5t² -15t

                                             0    - t   - 3

                                                   +t   -3

                                                     0   -6

Quotient = 4t² - 5t - 1

Remainder = -6

The remainder is constant it always be -6

P(3) = 4(3)³ - 17(3)² + 14(3) - 3

       = 4(27) - 17(9) + 42 - 3

       = 108 - 153 + 42 - 3

       = 150 - 156

      = -6

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Use a system of linear equations with two variables and two equations to solve. A number is 9 more than another number. Twice
Troyanec [42]

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Answer:

  (x, y) = (3, 12)

Step-by-step explanation:

Using the given variable definitions, we can write the equations ...

  y = x + 9

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Solving the second equation for y, we have ...

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4 0
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}.

4 0
3 years ago
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