22201+54633333333333

Consider a population p of field mice that grows at a rate proportional to the current population, so that dp dt = rp. (note: re

krok68 [10]

This is the concept of differential equations, given that
dp/dt=rp
then:
dp/p=rdt
thus:
ln p=rt+C
p=e^(rt+C)
P=Ke^(rt)
when t=210 days=7 months, p=2k
2k=k×e^(7r)
2=e^(7r)
ln2=7r
r=ln(2)/7
r=0.0990

5 0

3 years ago

It takes three hours for a boat to go 18 miles downstream, and nine hours to return. What is the rate of the boat in calm water

vampirchik [111]

Step-by-step explanation:

Distance = rate × time

If B is the rate of the boat, and C is the rate of the current, then we can write two equations:

18 = (B + C) × 3

18 = (B − C) × 9

Simplifying the equations:

6 = B + C

2 = B − C

Add the equations together to eliminate C:

8 = 2B

B = 4

Plug into either equation to find C:

C = 2

The boat's rate is 4 mph, and the current's rate is 2 mph.

5 0

3 years ago

every 5 1/2 cups of flour weighs 1 9/16 pounds. use a unit rate to show how u could determine if there are more than lee than 35

umka21 [38]

Okay. So to find the unit rate,divide pounds of flour by the number of cups. Turn the mixed numbers into improper fractions, which are 11/2 and 25/16. To divide fractions, keep the fraction the same, change the division symbol into multiplication, and flip the second fraction over. Then, multiply straight across. 11/2 * 16/25 is 176/50 or 3 13/25 in simplest form. To find out if there are more or less than 35 cups of flour in a 10-pound bag, you can round the number to the nearest whole number and multiply it by 10. For the exact unit rate, there are 3 13/25 cups of flour per 1 pound bag. Now, multiply that number by 10. Going back to improper fractions, 176/50 * 10/1 is 1,760/50 or 35 1/5 in simplest form. There are more than 35 cups of flour in a 10-pound bag of flour.

7 0

3 years ago

Anyone know the answer to this algebra problem?

ikadub [295]

Answer: $\bold{a=1\qquad b=\dfrac{1}{16}\qquad c=\dfrac{1}{64}\qquad d=1\qquad e=\dfrac{4}{9}\qquad f=\dfrac{16}{81}}$

<u>Step-by-step explanation:</u>

$\begin{array}{c|l}\underline{\quad x\quad}&\underline{\quad 4^{-x}\qquad \qquad}\\-1&4^{-(-1)}=4^1=4\\\\0&4^{-(0)}=4^0=1\\\\2&4^{-(2)}=\dfrac{1}{4^2}=\dfrac{1}{16}\\\\4&4^{-(4)}=\dfrac{1}{4^4}=\dfrac{1}{64}\end{array}$

$\begin{array}{c|l}\underline{\quad \bigg x \quad}&\underline{\quad \bigg(\dfrac{2}{3}\bigg)^x\qquad \qquad}\\\\-1&\bigg(\dfrac{2}{3}\bigg)^{-1}=\dfrac{3}{2}\\\\0&\bigg(\dfrac{2}{3}\bigg)^{0}=1\\\\2&\bigg(\dfrac{2}{3}\bigg)^{2}=\dfrac{2^2}{3^2}=\dfrac{4}{9}\\\\4&\bigg(\dfrac{2}{3}\bigg)^{4}=\dfrac{2^4}{3^4}=\dfrac{16}{81}\end{array}$

3 0

3 years ago

The landing restuarant order 200 flowers for mother day.the orderd carnation at 1.50$,Rose's at 5.75$and the daisies at 2.60$ ea

Vanyuwa [196]

Answer:

She ordered 80 carnations, 50 roses, and 70 daisies.

Step-by-step explanation:

Let x = number of carnations, y = number of roses, and z = number of daisies. This problem gives us three equations.

First, remember that we multiply the price of each flower by the number bought then add up the subtotals, that gives the total amount paid.

Therefore,

1.50x + 5.75y + 2.60z = 589.50

Second, the sum of the numbers purchased is the total bought. Therefore,

x + y + z = 200

Finally, there are 20 fewer roses than daisies. Therefore,

y = z - 20

If you substitute z - 20 in for y in the first two equations, you can rewrite them so they only have x and z.

1.50x + 5.75(z - 20) + 2.60z = 589.50

1.50x + 5.75z - 115 + 2.60z = 589.50

1.50x + 8.35z = 704.50

x + (z - 20) + z = 200

x + 2z - 20 = 200

x + 2z = 220

If you solve the rewrite second equation for x, you get x = 220 - 2z. Therefore, you can substitute 220 - 2z in for x in the rewritten first equation, and then solve for z.

1.50 (220 - 2z) + 8.35z = 704.50

330 - 3z + 8.35z = 704.50

330 + 5.35z = 704.50

5.35z = 374.50

z = 70

If you have z = 70, then x = 220 - 2(70) = 80 and y = 70 - 20 = 50.

She ordered 80 carnations, 50 roses, and 70 daisies.

*HOPE THIS HELPED!!! :D)

4 0

3 years ago