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3 years ago

Solve for k 4k - 6 1 3k - 9 = 3

Mathematics

2 answers:

spin [16.1K]3 years ago

7 0

Add nine to both sides then divide by 3 and K=4

zhuklara [117]3 years ago

6 0

→ Answer For the first equation:

$k = \frac{7}{4}$

→ Step 1: Add 6 to both the sides:

$4k-6+6=1+6$

$4k=7$

→ Step 2: Divide both sides by 4:

$\frac{4k}{4} =\frac{7}{4}$

$k=\frac{7}{4}$

_________________________________________________

→ Answer for the second equation:

$k=4$

→ Step 1: Add 9 to both sides:

$3k-9+9=3+9$

$3k=12$

→ Step 2: Divide both sides by 3:

$\frac{3k}{3} =\frac{12}{3}$

$k=4$

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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:

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

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

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

$P(\infty) = 1200$

5 0

3 years ago

Charlie wants to swim across a river that's is 400 meters wide. He begins swimming perpendicular to the shore he from but ends u

nirvana33 [79]

Let's calculate Charlie swimming distance. Assume Charlie swimming distance is D.
D^2 = 400^2 + 100^2 => D = 412 meters

Assume d is Carter's distance.
d^2 = 400^2 + 150^2 => d = 427 meters

Carter swims more than Charlie by:
d - D = 427 - 412 = 15 meters

4 0

3 years ago

Really need help on this guys!

Sergio039 [100]

<h2>Hello!</h2>

The answer is:

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To solve the problem, we need to use the given equation which is in the general form, and then, use it transform it to the standard form in order to find the center of the circle and its radius.

So,

We are given the circle:

$x^{2}+y^{2}+18x+4y+49=0$