Ask question

Ask question

All categories

Free_Kalibri [48]

3 years ago

14

A quadrilateral has vertices (2, 0), (0, –2), (–2, 4), and (–4, 2). Which special quadrilateral is formed by connecting the midp

oints of the sides?
a.kite
b.rectangle
c.trapezoid
d.rhombus

1 answer:

HACTEHA [7]3 years ago

8 0

In order to know what the polygon is, you have to plot the coordinates. After plotting, it is obviously shown a form of a rectangle. After connecting the midpoints of the sides, it formed a (D) rhombus, not a kite.

You might be interested in

How many grams are in 15 kilograms? 1.5 150 1,500 15,000

diamong [38]

15000 because 15kg x 1000 = 15000g

8 0

3 years ago

Read 2 more answers

The computers of six faculty members in a certain department are to be replaced. Two of the faculty members have selected laptop

Anettt [7]

Answer:

a. $\frac{1}{15}$

b. $\frac{2}{5}$

c. $\frac{14}{15}$

d. $\frac{8}{15}$

Step-by-step explanation:

Given that there are two laptop machines and four desktop machines.

On a day, 2 computers to be set up.

To find:

a. probability that both selected setups are for laptop computers?

b. probability that both selected setups are desktop machines?

c. probability that at least one selected setup is for a desktop computer?

d. probability that at least one computer of each type is chosen for setup?

Solution:

Formula for probability of an event E can be observed as:

$P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}$

a. Favorable cases for Both the laptops to be selected = $_2C_2$ = 1

Total number of cases = 15

Required probability is $\frac{1}{15}$ .

b. Favorable cases for both the desktop machines selected = $_4C_2=6$

Total number of cases = 15

Required probability is $\frac{6}{15} = \frac{2}{5}$ .

c. At least one desktop:

Two cases:

1. 1 desktop and 1 laptop:

Favorable cases = $_2C_1\times _4C_1 = 8$

2. Both desktop:

Favorable cases = $_4C_2=6$

Total number of favorable cases = 8 + 6 = 14

Required probability is $\frac{14}{15}$ .

d. 1 desktop and 1 laptop:

Favorable cases = $_2C_1\times _4C_1 = 8$

Total number of cases = 15

Required probability is $\frac{8}{15}$ .

8 0

3 years ago

What is the arc length of the subtending arc for an angle of 72 degrees on a circle of radius 4?

skad [1K]

Use a portion of the circumference: pi x 2 x r x angle at centre =8xpix72/360 =8pi/5

3 0

3 years ago

Calculus 2 Master needed, evaluate the indefinite integral of: <img src="https://tex.z-dn.net/?f=%5Cint%5C%28%20%28lnx%29%5E2%7D

viva [34]

Answer:

$\int (\ln(x))^2dx=x(\ln(x)^2-2\ln(x)+2)+C$

Step-by-step explanation:

So we have the indefinite integral:

$\int (\ln(x))^2dx$

This is the same thing as:

$=\int 1\cdot (\ln(x))^2dx$

So, let's do integration by parts.

Let u be (ln(x))². And let dv be (1)dx. Therefore:

$u=(\ln(x))^2\\\text{Find du. Use the chain rule.}\\\frac{du}{dx}=2(\ln(x))\cdot\frac{1}{x}$

Simplify:

$du=\frac{2\ln(x)}{x}dx$

And:

$dv=(1)dx\\v=x$

Therefore:

$\int (\ln(x))^2dx=x\ln(x)^2-\int(x)(\frac{2\ln(x)}{x})dx$

The x cancel:

$=x\ln(x)^2-\int2\ln(x)dx$

Move the 2 to the front:

$=x\ln(x)^2-2\int\ln(x)dx$

(I'm not exactly sure how you got what you got. Perhaps you differentiated incorrectly?)

Now, let's use integrations by parts again for the integral. Similarly, let's put a 1 in front:

$=x\ln(x)^2-2\int 1\cdot\ln(x)dx$

Let u be ln(x) and let dv be (1)dx. Thus:

$u=\ln(x)\\du=\frac{1}{x}dx$

And:

$dv=(1)dx\\v=x$

So:

$=x\ln(x)^2-2(x\ln(x)-\int (x)\frac{1}{x}dx)$

Simplify the integral:

$=x\ln(x)^2-2(x\ln(x)-\int (1)dx)$

Evaluate:

$=x\ln(x)^2-2(x\ln(x)-x)$

Now, we just have to simplify :)

Distribute the -2:

$=x\ln(x)^2-2x\ln(x)+2x$

And if preferred, we can factor out a x:

$=x(\ln(x)^2-2\ln(x)+2)$

And, of course, don't forget about the constant of integration!

$=x(\ln(x)^2-2\ln(x)+2)+C$

And we are done :)

8 0

3 years ago

A ball is dropped from a height of 10 feet and returns to a height that is one-half of the height from which it fell. The ball c

ANTONII [103]

1st drop- 10ft comes back up 5ft
2nd drop-5ft comes back up 2.5ft
3rd drop-2.5ft comes back up 1.25 ft
4th drop-1.25ft comes back up .625ft

(10+5) + (5+2.5) + (2.5+1.25) + (1.25) = 27.5ft

7 0

3 years ago