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3 years ago

What is the simplified fractional equivalent of the terminating decimal 0.75

Mathematics

1 answer:

Ymorist [56]3 years ago

3 0

0.75 (decimal) is equal to 3/4 (simplified fraction)

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Subtract (6k + 2) - (3k - 2)

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The answer is (3k + 4)

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Find the OPPOSITE of each integer. -3

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Step-by-step explanation:

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How do you define each step of this -3(y+3)=2y+3

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$-3(y+3)=2y+3
\\\ Distributation
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3 years ago

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

trasher [3.6K]

Answer:

(a) 100 fishes

(b) t = 10: 483 fishes

t = 20: 999 fishes

t = 30: 1168 fishes

(c)

$P(\infty) = 1200$

Step-by-step explanation:

Given

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

$d = 1200\\k = 11\\c=0.2$

Solving (a): Fishes at t = 0

This gives:

$P(0) =\frac{1200}{1+11*e^-{0.2*0}}$

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

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

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

$P(0) =\frac{1200}{12}$

$P(0) = 100$

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

$t = 10$

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

$t = 20$

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

$t = 30$

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

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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3 years ago

A mixture weighing 600kg has two ingredients: sand, cement and stones. There is twice as much cement as sand and three times as

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Answer:

Step-by-step explanation:

All the masses here are based on the main mass of the sand. So we will call sand "x". If there is two times as much cement as sand, then cement is 2x; if there is three times as much stone as sand, then stone is 3x. All of these added together equal a mass of 600 kg:

x + 2x + 3x = 600 so

6x = 600 and

x = 100 kg

There are 100 kg of sand. That means that there is 200 kg of cement and 300 kg of stone

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