All categories

2 years ago

1. {(-1, -9), (0, -5), (4,3), (6, -2)} D= R = _ Function?

Mathematics

1 answer:

Studentka2010 [4]2 years ago

7 0

Answer:

D=(-1,0,4,6)

R=(-9,-5,-2,3)

Function

You might be interested in

Snow is falling steadily in Syracuse, New York. After 2 hours, 4 inches of snow has fallen

zysi [14]

Answer:

Step-by-step explanation:

2 hours = 4 inches

2/2 equals 1 hour

4/2 equals 2 inches

2 inches of snow per hour

6.5*2 is 13

you should expect 13 inches of snow in 6.5 hours

4 0

3 years ago

Market-share-analysis company Net Applications monitors and reports on Internet browser usage. According to Net Applications, in

EastWind [94]

The answer is c. I believe so

4 0

3 years ago

Rewa Delta Union RugbyCEOhas become concerned about the slow pace of the rugby gamesplayed inthe current union rugby, fearing th

Alex Ar [27]

Answer:

Check the explanation

Step-by-step explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

8 0

2 years ago

The figure shows two intersecting lines and the measures of the resulting angles. Write an equation to help you solve for x. The

Svetlanka [38]

Given:

The given figure shows the intersection of the two lines.

The angles formed by the intersection of the two lines are (3x - 8)° and (2x + 12)°

We need to determine the equation to solve for x and to find the value of x.

Equation to solve for x:

Since, the two angles (3x - 8)° and (2x + 12)° are vertically opposite angles and the vertical angles are always equal.

Hence, we have;

$3x-8=2x+12$

Thus, the equation to solve for x is $3x-8=2x+12$

Value of x:

The value of x can be determined by solving the equation $3x-8=2x+12$

Thus, we have;

$x-8=12$

$x=20$

Thus, the value of x is 20.

3 0

3 years ago

How do you integrate arctan(x dx? i think that if you simplify the integral you get:?

Temka [501]

Integration by parts will help here. Letting $u=\arctan x$ and $\mathrm dv=\mathrm dx$ , you end up with $\mathrm du=\dfrac{\mathrm dx}{1+x^2}$ and $v=x$ . Now

$\displaystyle\int\arctan x\,\mathrm dx=uv-\int v\,\mathrm du$
$\displaystyle\int\arctan x\,\mathrm dx=x\arctan x-\int\frac x{1+x^2}\,\mathrm dx$

For the remaining integral, setting $y=1+x^2$ gives $\dfrac{\mathrm dy}2=x\,\mathrm dx$ , so

$\displaystyle\int\frac x{1+x^2}\,\mathrm dx=\frac12\int\frac{\mathrm dy}y=\frac12\ln|y|+C=\frac12\ln(1+x^2)+C$

Putting everything together, you end up with

$\displaystyle\int\arctan x\,\mathrm dx=x\arctan x-\frac12\ln(1+x^2)+C$

6 0

2 years ago