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IgorC [24]
3 years ago
5

What is the coefficent of the term 6xy/7

Mathematics
2 answers:
seraphim [82]3 years ago
8 0
Simple...

\frac{6xy}{7}

Its Leading Coefficient is \frac{6}{7}

Its Polynomial Degree is 2

Its Leading Term \frac{6xy}{7}

Thus, your answer.
chubhunter [2.5K]3 years ago
4 0
The coefficient is the number 6/7.
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A wildlife photographer photographs a crocodile and then paddles against a river current for 6 km. Then, hoping to photograph a
Angelina_Jolie [31]

Answer:

5 or 6

Step-by-step explanation:

So I did it in a different way than shown in rsm classes. I said that the speed was x-2 and x.

then I said the distance was 6 and 15. If the time taken was 1 hour more, then we can come up with this equation:

Since the time is 6/x-2 or 15/x

15/x-6/x-2=1

multiply everything to get a common denominator then cancel it out to get

(15x-30)-6x=x^2-2x

11x-30=x^2

x^2-11x+30=0

(x-5)(x-6) is factored form.

so x=5 or x=6. this means that the speed of the boat is either 5km/h or 6 km/h

6 0
2 years ago
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3 years ago
2 0,484,163 expanded form
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20,000,000+400,000+80,000+4,000+100+60+3
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3 years ago
A lake polluted by bacteria is treated with an antibacterial chemical. Aftertdays, thenumber N of bacteria per milliliter of wat
jeyben [28]

Answer:

N(1)=50 is a minimum

N(15)=4391.7 is a maximum

Step-by-step explanation:

<u>Extrema values of functions </u>

If the first and second derivative of a function f exists, then f'(a)=0 will produce values for a called critical points. If a is a critical point and f''(a) is negative, then x=a is a local maximum, if f''(a) is positive, then x=a is a local minimum.  

We are given a function (corrected)

N(t) = 20(t^2-lnt^2)+ 30

N(t) = 20(t^2-2lnt)+ 30

(a)

First, we take its derivative

N'(t) = 20(2t-\frac{2}{t})

Solve N'(t)=0

20(2t-\frac{2}{t})=0

Simplifying

2t^2-2=0

Solving for t

t=1\ ,t=-1

Only t=1 belongs to the valid interval 1\leqslant t\leqslant 15

Taking the second derivative

N''(t) = 20(2+\frac{2}{t^2})

Which is always positive, so t=1 is a minimum

(b)

N(1)=20(1^2-2ln1)+ 30

N(1)=50 is a minimum

(c) Since no local maximum can be found, we test for the endpoints. t=1 was already determined as a minimum, we take t=15

(d)

N(15)=20(15^2-2ln15)+ 30

N(15)=4391.7 is a maximum

7 0
3 years ago
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