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2 years ago

The table below shows the data for a random sample of fish that were collected from and later released into a lake:

Mathematics

2 answers:

Lubov Fominskaja [6]2 years ago

6 0

The answer:
It is pretty sure 244

Olin [163]2 years ago

5 0

Answer:

244

Step-by-step explanation:

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Samantha is using a 2-liter pitcher to serve lemonade to 10 of her friends. How many times will she need to fill the pitcher in

Mice21 [21]

Answer: 2 times

Step-by-step explanation:

Hi, since 1000 milliliters = 1 liter

We have to convert the 2 liters into milliliters:

2 x 1000 = 2000 milliliters, so she is using a 2,000 milliliters pincher.

Now we have to multiply the number of friends (10) by the amount of lemonade she serves to each one (400 ml)

400 x 10 = 4,000 milliliters

Finally, we have to divide the amount of lemonade she needs to serve (4,000 ml) by the capacity of the pitcher (2,000), to obtain the number of times that she will need to fill the pitcher.

4,000/2,000=2 times

Feel free to ask for more if needed or if you did not understand something.

4 0

3 years ago

Read 2 more answers

A = ?<br> b = ?<br> Please explain as well :)

sweet [91]

Answer:

a = 25 , b = 14

Step-by-step explanation:

For A

a/5 + 3 = 8

a/5 = 8 - 3

a/5 = 5

Cross multiply

a = 5 * 5

a = 25

For B

3b/7 - 1 = 5

3b/7 = 5 + 1

3b/7 = 6

Cross multiply

3b = 42

b = 42 / 3

b = 14

Hope it will help

5 0

3 years ago

Find the function y1 of t which is the solution of 121y′′+110y′−24y=0 with initial conditions y1(0)=1,y′1(0)=0. y1= Note: y1 is

strojnjashka [21]

Answer:

Step-by-step explanation:

The original equation is $121y''+110y'-24y=0$ . We propose that the solution of this equations is of the form $y = Ae^{rt}$ . Then, by replacing the derivatives we get the following

$121r^2Ae^{rt}+110rAe^{rt}-24Ae^{rt}=0= Ae^{rt}(121r^2+110r-24)$

Since we want a non trival solution, it must happen that A is different from zero. Also, the exponential function is always positive, then it must happen that

$121r^2+110r-24=0$

Recall that the roots of a polynomial of the form $ax^2+bx+c$ are given by the formula

$x = \frac{-b \pm \sqrt[]{b^2-4ac}}{2a}$

In our case a = 121, b = 110 and c = -24. Using the formula we get the solutions

$r_1 = -\frac{12}{11}$

$r_2 = \frac{2}{11}$

So, in this case, the general solution is $y = c_1 e^{\frac{-12t}{11}} + c_2 e^{\frac{2t}{11}}$

a) In the first case, we are given that y(0) = 1 and y'(0) = 0. By differentiating the general solution and replacing t by 0 we get the equations

$c_1 + c_2 = 1$

$c_1\frac{-12}{11} + c_2\frac{2}{11} = 0$ (or equivalently $c_2 = 6c_1$

By replacing the second equation in the first one, we get $7c_1 = 1$ which implies that $c_1 = \frac{1}{7}, c_2 = \frac{6}{7}$ .

So $y_1 = \frac{1}{7}e^{\frac{-12t}{11}} + \frac{6}{7}e^{\frac{2t}{11}}$

b) By using y(0) =0 and y'(0)=1 we get the equations

$c_1+c_2 =0$

$c_1\frac{-12}{11} + c_2\frac{2}{11} = 1$ (or equivalently $-12c_1+2c_2 = 11$

By solving this system, the solution is $c_1 = \frac{-11}{14}, c_2 = \frac{11}{14}$

Then $y_2 = \frac{-11}{14}e^{\frac{-12t}{11}} + \frac{11}{14} e^{\frac{2t}{11}}$

The Wronskian of the solutions is calculated as the determinant of the following matrix

$\left| \begin{matrix}y_1 & y_2 \\ y_1' & y_2'\end{matrix}\right|= W(t) = y_1\cdot y_2'-y_1'y_2$

By plugging the values of $y_1$ and

We can check this by using Abel's theorem. Given a second degree differential equation of the form y''+p(x)y'+q(x)y the wronskian is given by

$e^{\int -p(x) dx}$

In this case, by dividing the equation by 121 we get that p(x) = 10/11. So the wronskian is

$e^{\int -\frac{10}{11} dx} = e^{\frac{-10x}{11}}$

Note that this function is always positive, and thus, never zero. So $y_1, y_2$ is a fundamental set of solutions.

8 0

3 years ago

Find the surface area of the cylinder in terms of pi 14cm 18cm

Artist 52 [7]

350 cm^2 (A for connections academy)

3 0

3 years ago

How do i write 3% as a fraction

Ahat [919]

3/10
As a fraction bababababa

4 0

3 years ago