In how many ways can you select 3 books to bring on vacation from a group of 11 books?

The basketball team practices every day from 2:30 pm to 4:00 pm.todays practice ended 20 minutes early.at what time did basketba

Goshia [24]

3:40 because 4:00 minus 20 is 3:40

8 0

3 years ago

A bag contains 5 red marbles, two orange marbles, one yellow marbles, and two green marbles. Two marbles are drawn from the bag.

Mandarinka [93]

Green marble- probability is 2 out of 10
orange-1 out of 10

5 0

2 years ago

Read 2 more answers

What is the constant of proportionality in the equation X/y=2/9

mart [117]

Answer:

the answer is 2/9

Step-by-step explanation:

x/y=2/9

y=9/2 x

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

c)

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

2 years ago

A certain region currently has wind farms capable of generating a total of 2200 megawatts (2.2 gigawatts) of power. Assuming win

sveta [45]

Answer:

The correct answer is A. 4,818'000,000 kilowatt-hours per year and B. 481,800 households.

Step-by-step explanation:

1. Let's review the information provided to us for solving the questions:

Power capacity of the wind farms = 2,200 Megawatts or 2.2 Gigawatts

2. Let's resolve the questions A and B:

Part A

Assuming wind farms typically generate 25% of their capacity, how much energy, in kilowatt-hours, can the region's wind farms generate in one year?

2,200 * 0.25 = 550 Megawatts

550 Megawatts = 550 * 1,000 Kilowatts = 550,000 Kilowatts

Now we calculate the amount of Kilowatts per hour, per day and per year:

550,000 Kw generated by the farms means that are capable of produce 550,000 kw per hour of energy

550,000 * 24 = 13'200,000 kilowatt-hours per day

13'200,000 * 365 = 4,818'000,000 kilowatt-hours per year

Part B

Given that the average household in the region uses about 10,000 kilowatt-hours of energy each year, how many households can be powered by these wind farms?

For calculating the amount of households we divide the total amount of energy the wind farms can generate (4,818'000,000 kilowatt-hours) and we divide it by the average household consumption (10,000 kilowatt-hours)

Amount of households = 4,818'000,000/10,000 = 481,800

8 0

3 years ago