you can put in any of the numbers that are already listed in the x values. You can put 5, 14, -1, and 23 because something can't be a function if x values repeat.

8 0

2 years ago

The coordinates of the vertices of a rectangle are (−3, 4) , (7, 2) , (6, −3) , and (−4, −1) . What is the perimeter of the rect

ki77a [65]

Answrer

Find out the what is the perimeter of the rectangle .

To prove

Now as shown in the figure.

Name the coordinates as.

A(−3, 4) ,B (7, 2) , C(6, −3) , and D(−4, −1) .

In rectangle opposite sides are equal.

Thus

AB = DC

AD = BC

Formula

$Disatnce\ formula = \sqrt{(x_{2} - x_{1})^{2} +(y_{2} - y_{1})^{2}}$

Now the points A(−3, 4) and B(7, 2)

$AB = \sqrt{(7- (-3))^{2} +(2- 4)^{2}}$

$AB = \sqrt{(10)^{2} +(-2)^{2}}$

$AB = \sqrt{100+4}$

$AB = \sqrt{104}$

$AB = 2\sqrt{26}\units$

Thus

$CD= 2\sqrt{26}\units$

Now the points

A (−3, 4) , D (−4, −1)

$AD = \sqrt{(-4 - (-3))^{2} +(-1- 4)^{2}}$

$AD = \sqrt{(-1)^{2} +(-5)^{2}}$

$AD = \sqrt{1 + 25}$

$AD = \sqrt{26}\units$

Thus

$BC = \sqrt{26}\units$

Formula

Perimeter of rectangle = 2 (Length + Breadth)

Here

$Length = 2\sqrt{26}\ units$

$Breadth = \sqrt{26}\ units$

$Perimeter\ of\ rectangle = 2(2\sqrt{26} +\sqrt{26})$

$Perimeter\ of\ rectangle = 2(3\sqrt{26})$

$Perimeter\ of\ rectangle = 6\sqrt{26}$

$\sqrt{26} = 5.1 (Approx)$

$Perimeter\ of\ rectangle = 6\times 5.1$

Perimeter of a rectangle = 30.6 units.

Therefore the perimeter of a rectangle is 30.6 units.

8 0

2 years ago

An English teacher has 6 short stories, 4 novels, and 23 poems to choose from. How many ways can he assign one of each to his cl

Karolina [17]

Answer:

552

Step-by-step explanation:

This is a problem of permutation which can be solved by rule of fundamental counting principle.

This principle states that if there "m" ways of doing one thing and "n" ways of doing other. Then no. of ways in which both the things can be done together is "m*n". This can be extended for m, n, p,r, s things and so on.

example: if there are 5 shirts and 3 trousers then number of ways in which the shirts and trousers can be worn is 5*3 = 15 ways.

_____________________________________________

The given problem is on similar concepts.

here 6 short stories, 4 novels, and 23 poems have to be assigned to his class.

Thus it can be done in 6*4*23 = 552 ways.

5 0

2 years ago