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victus00 [196]
2 years ago
7

Write y=(-3/4)x+3 in standard form using integers

Mathematics
1 answer:
aniked [119]2 years ago
7 0

Answer:

Step-by-step explanation:

multiply by 4 :

4y = - 3x + 12

3x + 4y = 12 is the  standard form using integers

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Population Growth A lake is stocked with 500 fish, and their population increases according to the logistic curve where t is mea
Alexus [3.1K]

Answer:

a) Figure attached

b) For this case we just need to see what is the value of the function when x tnd to infinity. As we can see in our original function if x goes to infinity out function tend to 1000 and thats our limiting size.

c) p'(t) =\frac{19000 e^{-\frac{t}{5}}}{5 (1+19e^{-\frac{t}{5}})^2}

And if we find the derivate when t=1 we got this:

p'(t=1) =\frac{38000 e^{-\frac{1}{5}}}{(1+19e^{-\frac{1}{5}})^2}=113.506 \approx 114

And if we replace t=10 we got:

p'(t=10) =\frac{38000 e^{-\frac{10}{5}}}{(1+19e^{-\frac{10}{5}})^2}=403.204 \approx 404

d) 0 = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}

And then:

0 = 7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)

0 =19e^{-\frac{t}{5}} -1

ln(\frac{1}{19}) = -\frac{t}{5}

t = -5 ln (\frac{1}{19}) =14.722

Step-by-step explanation:

Assuming this complete problem: "A lake is stocked with 500 fish, and the population increases according to the logistic curve p(t) = 10000 / 1 + 19e^-t/5 where t is measured in months. (a) Use a graphing utility to graph the function. (b) What is the limiting size of the fish population? (c) At what rates is the fish population changing at the end of 1 month and at the end of 10 months? (d) After how many months is the population increasing most rapidly?"

Solution to the problem

We have the following function

P(t)=\frac{10000}{1 +19e^{-\frac{t}{5}}}

(a) Use a graphing utility to graph the function.

If we use desmos we got the figure attached.

(b) What is the limiting size of the fish population?

For this case we just need to see what is the value of the function when x tnd to infinity. As we can see in our original function if x goes to infinity out function tend to 1000 and thats our limiting size.

(c) At what rates is the fish population changing at the end of 1 month and at the end of 10 months?

For this case we need to calculate the derivate of the function. And we need to use the derivate of a quotient and we got this:

p'(t) = \frac{0 - 10000 *(-\frac{19}{5}) e^{-\frac{t}{5}}}{(1+e^{-\frac{t}{5}})^2}

And if we simplify we got this:

p'(t) =\frac{19000 e^{-\frac{t}{5}}}{5 (1+19e^{-\frac{t}{5}})^2}

And if we simplify we got:

p'(t) =\frac{38000 e^{-\frac{t}{5}}}{(1+19e^{-\frac{t}{5}})^2}

And if we find the derivate when t=1 we got this:

p'(t=1) =\frac{38000 e^{-\frac{1}{5}}}{(1+19e^{-\frac{1}{5}})^2}=113.506 \approx 114

And if we replace t=10 we got:

p'(t=10) =\frac{38000 e^{-\frac{10}{5}}}{(1+19e^{-\frac{10}{5}})^2}=403.204 \approx 404

(d) After how many months is the population increasing most rapidly?

For this case we need to find the second derivate, set equal to 0 and then solve for t. The second derivate is given by:

p''(t) = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}

And if we set equal to 0 we got:

0 = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}

And then:

0 = 7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)

0 =19e^{-\frac{t}{5}} -1

ln(\frac{1}{19}) = -\frac{t}{5}

t = -5 ln (\frac{1}{19}) =14.722

7 0
2 years ago
The multimeter read 0.0425 volts, 0.1582 volts and 0.0932 volts. Which is the highest reading on the multimeter?
vladimir1956 [14]

Answer:

any number closest to 1 on the number line will be the highest reading.

.1582 is the highest number

Step-by-step explanation:


6 0
2 years ago
Try factoring this before moving on: 25+10x+x^2
Naddika [18.5K]

Answer:

Step-by-step explanation:

x2 + 10x + 25

x2 + 5x + 5x + 25

x (x + 5) + 5(x + 5)

(x + 5) (x + 5)

4 0
3 years ago
The granola summer buys used to cost $6.00 per pound, but it has been marked up 15%. How much in dollars and cents will summer p
Norma-Jean [14]
Summer will pay $17.94 for 2.6 lbs of granola

6 0
2 years ago
Jerald jumped from a bungee tower. If the equation that models his height, in feet, is h = –16t2 + 729, where t is the time in s
Snezhnost [94]

Answer. First option: t > 6.25


Solution:

Height (in feet): h=-16t^2+729

For which interval of time is h less than 104 feet above the ground?

h < 104

Replacing h for -16t^2+729

-16t^2+729 < 104

Solving for h: Subtracting 729 both sides of the inequality:

-16t^2+729-729 < 104-729

-16t^2 < -625

Multiplying the inequality by -1:

(-1)(-16t^2 < -625)

16t^2 > 625

Dividing both sides of the inequality by 16:

16t^2/16 > 625/16

t^2 > 39.0625

Replacing t^2 by [ Absolute value (t) ]^2:

[ Absolute value (t) ]^2 > 39.0625

Square root both sides of the inequality:

sqrt { [ Absolute value (t) ]^2 } > sqrt (39.0625)

Absolute value (t) > 6.25

t < -6.25 or t > 6.25, but t can not be negative, then the solution is:

t > 6.25



5 0
3 years ago
Read 2 more answers
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