B. The statement is true because if the degree of the numerator of a rational function equals the degree of the denominator, then the rational function has a horizontal asymptote that is equal to the ratio of the leading coefficients.
Collect like terms
-2x is the answer
Yes, it is. You may prove it using something like this: 1 is an integer. 1-1 is a difference between integers. A difference between integers returns an integer, so 0 is an integer.

- Given - <u>A </u><u>trapezium</u><u> </u><u>ABCD </u><u>with </u><u>non </u><u>parallel </u><u>sides </u><u>of </u><u>measure </u><u>1</u><u>5</u><u> </u><u>cm </u><u>each </u><u>!</u><u> </u><u>along </u><u>,</u><u> </u><u>the </u><u>parallel </u><u>sides </u><u>are </u><u>of </u><u>measure </u><u>1</u><u>3</u><u> </u><u>cm </u><u>and </u><u>2</u><u>5</u><u> </u><u>cm</u>
- To find - <u>Area </u><u>of </u><u>trapezium</u>
Refer the figure attached ~
In the given figure ,
AB = 25 cm
BC = AD = 15 cm
CD = 13 cm
<u>Construction</u><u> </u><u>-</u>

Now , we can clearly see that AECD is a parallelogram !
AE = CD = 13 cm
Now ,

Now , In ∆ BCE ,

Now , by Heron's formula

Also ,

<u>Since </u><u>we've </u><u>obtained </u><u>the </u><u>height </u><u>now </u><u>,</u><u> </u><u>we </u><u>can </u><u>easily </u><u>find </u><u>out </u><u>the </u><u>area </u><u>of </u><u>trapezium </u><u>!</u>

hope helpful :D
The line is starting from right side and goes thorough left .
Means
- Its starts from 1st Quadrant and goes to 3rd Quadrant.(+ve to -ve)
- Y is infinite
Hence
The solution is
![\\ \sf\longmapsto (0,\infty]](https://tex.z-dn.net/?f=%5C%5C%20%5Csf%5Clongmapsto%20%280%2C%5Cinfty%5D)