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2 years ago

300 x 25.4 SHOW YOUR WORK

Mathematics

1 answer:

Lana71 [14]2 years ago

6 0

Answer:

7620

Step-by-step explanation:

25x 3 = 75 so 25x 300 = 7500

4x 300 = 1200

0.4x300 = 120

7500+120 = 7620

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Brainliest for whoever gets this right!

My name is Ann [436]

Answer:

$x=\frac{(a^2+ab)}{a-b}$

Step-by-step explanation:

So we want to make x the subject of the formula:

$a(x-b)=a^2+bx$

First, distribute the left side:

$a(x)-a(b)=a^2+bx\\ax-ab=a^2+bx$

Combine all the terms with x with it on one side. To do this, first subtract bx from both sides. The right side cancels:

$(ax-ab)-bx=(a^2+bx)-bx\\ax-ab-bx=a^2$

Remove the -ab from the left. Add ab to both sides. The left side cancels:

$(ax-ab-bx)+ab=a^2+ab\\ax-bx=a^2+ab$

Now, distribute out the x from the left side:

$ax-bx=a^2+ab\\x(a-b)=a^2+ab$

Divide both sides by (a-b). The left side cancels:

$\frac{(x(a-b))}{a-b}=\frac{(a^2+ab)}{a-b} \\x=\frac{(a^2+ab)}{a-b}$

Therefore:

$x=\frac{(a^2+ab)}{a-b}$

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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