Match the function with its graph.
1) y= tan x
2) y= cot x
3) y= -tan x
4) y= -cot x

Option a. 1C, 2A, 3B, 4D.

Step-by-step explanation:

1) We know that tan(x)=sin(x)/cos(x). If x=0, sin(x)=0 and cos(x)=1 then tan(x)=0. For that reason, we know that the graph passes through the point (0,0).

If x=45, then sin(45)= and cos(45)=. Thus tan(45)=1. The only graph that passes through the point (0,0) and is possitive when x=45 is the graph C.

2) We know that cot(x)=cos(x)/sin(x). If x=0, sin(x)=0 and cos(x)=1 then tan(x)=+∞. For that reason, we know that the graph has an asymptote in y=0, in other words, it never crosses the y-axis.

If x=45, then sin(45)= and cos(45)=. Thus cot(45)=1. The only graph that has an asymptote in y=0 and is possitive when x=45 is the graph A.

3) We know that -tan(x)=-sin(x)/cos(x). If x=0, sin(x)=0 and cos(x)=1 then -tan(x)=0. For that reason, we know that the graph passes through the point (0,0).

If x=45, then sin(45)= and cos(45)=. Thus -tan(45)=-1. The only graph that passes through the point (0,0) and is negative when x=45 is the graph B.

) We know that -cot(x)=-cos(x)/sin(x). If x=0, sin(x)=0 and cos(x)=1 then tan(x)=-∞. For that reason, we know that the graph has an asymptote in y=0, in other words, it never crosses the y-axis.

If x=45, then sin(45)= and cos(45)=. Thus -cot(45)=-1. The only graph that has an asymptote in y=0 and is negative when x=45 is the graph D.

is there answer choices?

x = 2

step-by-step explanation:

if sr parallel to ba then the sides bc and ac are in proportion, that is

= ← substitute values

= ( cross- multiply )

27x = 6(4x + 1)

27x = 24x + 6 ( subtract 24x from both sides )

3x = 6 ( divide both sides by 3 )

x = 2