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3 years ago

Which equation represents a line that passes through (–9, –3) and has a slope of –6?

Mathematics

2 answers:

Helen [10]3 years ago

3 0

Answer:

y+3= -6(x+9) is the answer

Step-by-step explanation:

dem82 [27]3 years ago

3 0

Answer:

$y+3=-6(x+9)$

Step-by-step explanation:

We are given that

Slope of a line=-6

Given point =(-9,-3)

We have to find the equation which represents the line.

The equation of line passing through the given point $(x_1,y_1)$ with slope m is given by

$y-y_1=m(x-x_1)$

Substitute the values then we get

The equation of line passing through the point (-9,-3) with slope -6 is given by

$y-(-3)=-6(x-(-9))$

$y+3=-6(x+9)$

Hence, the equation of line that passes through (-9,-3) and has a slope -6 is given by

$y+3=-6(x+9)$

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2 years ago

Animal populations are not capable of unrestricted growth because of limited habitat and food supplies. Under such conditions th

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(a) 100 fishes

(b) t = 10: 483 fishes

t = 20: 999 fishes

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(c)

$P(\infty) = 1200$

Step-by-step explanation:

Given

$P(t) =\frac{d}{1+ke^-{ct}}$

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Solving (a): Fishes at t = 0

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$P(0) =\frac{1200}{1+11*1}$

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Solving (a): Fishes at t = 10, 20, 30

$t = 10$

$P(10) =\frac{1200}{1+11*e^-{0.2*10}} =\frac{1200}{1+11*e^-{2}}\\\\P(10) =\frac{1200}{1+11*0.135}=\frac{1200}{2.485}\\\\P(10) =483$

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$P(20) =\frac{1200}{1+11*e^-{0.2*20}} =\frac{1200}{1+11*e^-{4}}\\\\P(20) =\frac{1200}{1+11*0.0183}=\frac{1200}{1.2013}\\\\P(20) =999$

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$P(30) =\frac{1200}{1+11*e^-{0.2*30}} =\frac{1200}{1+11*e^-{6}}\\\\P(30) =\frac{1200}{1+11*0.00247}=\frac{1200}{1.0273}\\\\P(30) =1168$

Solving (c): $\lim_{t \to \infty} P(t)$

In (b) above.

Notice that as t increases from 10 to 20 to 30, the values of $e^{-ct}$ decreases

This implies that:

${t \to \infty} = {e^{-ct} \to 0}$

So:

The value of P(t) for large values is:

$P(\infty) = \frac{1200}{1 + 11 * 0}$

$P(\infty) = \frac{1200}{1 + 0}$

$P(\infty) = \frac{1200}{1}$

$P(\infty) = 1200$

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2 years ago

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Answer:

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Since 1 Hour = 1 Dollar,

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2 years ago

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3 years ago

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Answer:

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