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2 years ago

Solve using square root method

Mathematics

1 answer:

avanturin [10]2 years ago

6 0

$\frac{1}{2}(x+5)^2 + 9 = -15$

$\frac{(x+5)^2+18}{2} = \frac{-30}{2}$

Cancel out the denominators.

(x+5)² + 18 = -30

(x+5)² = -30 - 18

(x+5)² = -48

The equation is impossible, we have a negative number.

There isn't a number that squared gives a negative number (the square is always positive).

√(x+5)² = +/- √-48

x+5 = +/- √-48

IMPOSSIBLE

You can write the result in imaginary numbers

x+5=+/- 4i√3

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432 people on Friday and 658 on Saturday

Step-by-step explanation:

Lets use variables for each day

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put them together

6x+8y=7856

x+y=1090

now you can cancel a variable but manipulating one of the equations. We'll use x, to cancel x you need make it zero so multiply the bottom by -6

6x+8y=7856

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(2y/2)=(1316/2)

y=658

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3 years ago

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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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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

$t = 10$

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

$t = 20$

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

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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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