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Vinvika [58]
3 years ago
13

In your notebook, set up the following subtraction in a vertical format and select the correct answer. Find 2p 2 3p - 4 less - 2

p 2 - 3p 4. 4p2 6p 8 4p2 - 6p - 8 4p2 6p - 8 0.
Mathematics
1 answer:
Elena L [17]3 years ago
3 0

The equation in the reduced form of the equation is \rm 4p^ 2 + 6p - 8.

Given that,

Equation; \rm  2p^ 2 + 3p - 4\  by\  -2p ^2 - 3p +4

We have to reduce the equation.

According to the question,

To reduce the equation means we need to subtract one equation from the other.

To determine the reduced form of the equation following all the steps given below.

Equation; \rm  2p^ 2 + 3p - 4\  by\  -2p ^2 - 3p +4

Subtraction equation 1 from equation 2,

\rm  = \rm  2p^ 2 + 3p - 4\  -(-2p ^2 - 3p + 4)\\\\=  2p^ 2 + 3p - 4\  -(-2p ^2 -3p +4)\\\\ = 2p^ 2 + 3p - 4+2p^ 2 + 3p - 4 \\\\ = 4p^ 2 + 6p - 8

Hence, The equation in the reduced form of the equation is \rm 4p^ 2 + 6p - 8.

For more details refer to the link given below.

brainly.com/question/1807316

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The ratio of boys to girls in a class is 6:4. If there is a total of 30 students in the class, how many are boys?
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Read 2 more answers
The length of some fish are modeled by a von Bertalanffy growth function. For Pacific halibut, this function has the form L(t) =
Dafna1 [17]

Answer:

a) L'(t) = 34.416*e^(-0.18*t)

b) L'(0) = 34 cm/yr , L'(1) =29 cm/yr , L'(6) =12 cm/yr

c) t = 10 year                                          

Step-by-step explanation:

Given:

- The length of fish grows with time. It is modeled by the relation:

                                   L(t) = 200*(1-0.956*e^(-0.18*t))

Where,

L: Is length in centimeter of a fist

t: Is the age of the fish in years.

Find:

(a) Find the rate of change of the length as a function of time

(b) In this part, give you answer to the nearest unit. At what rate is the fish growing at age: t = 0 , t = 1, t = 6

c) When will the fish be growing at a rate of 6 cm/yr? (nearest unit)

Solution:

- The rate of change of length of a fish as it ages each year  can be evaluated by taking a derivative of the Length L(t) function with respect to x. As follows:

                             dL(t)/dt = d(200*(1-0.956*e^(-0.18*t))) / dt

                             dL(t)/dt = 34.416*e^(-0.18*t)

- Then use the above relation to compute:

                            L'(t) = 34.416*e^(-0.18*t)

                            L'(0) = 34.416*e^(-0.18*0) = 34 cm/yr

                            L'(1) = 34.416*e^(-0.18*1) = 29 cm/yr

                            L'(6) = 34.416*e^(-0.18*6) = 12 cm/yr

- Next, again use the derived L'(t) to determine the year when fish is growing at a rate of 6 cm/yr:

                             6 cm/yr = 34.416*e^(-0.18*t)

                             e^(0.18*t) = 34.416 / 6

                             0.18*t = Ln(34.416/6)

                             t = Ln(34.416/6) / 0.18

                             t = 10 year

7 0
3 years ago
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