Informally, the domain is the set of all possible elements in a set for which there is one and only one output. In interval notation, the domain is (-4, ∞). I am assuming it continues on infinitely to the right because of the arrow. The range is [1, <span>∞). Here, you choose the lowest value on the graph, up to the highest one.</span>
In a table it's EXAMPLE: Cars/Drivers the cars is x and the drivers is y(y-intercept). In an equation, EXAMPLE using y=mx+b the b is the y-int., and in a graph it is (x,y) the y being the y-int.
Answer:
A: The interquartile range is 8.
Step-by-step explanation:
Hopefully this helps!
Answer: 195.5 or 12 7/32
Step-by-step explanation:
There is no letter tetha in the table so I use α instead. However it is not sence to final result.
The expression is:
(sinα+cosα)/(cosα*(1-cosα))
Lets divide the nominator and denominator by cosα
(sinα/cosα+cosα/cosα)/(cosα*(1-cosα)/cosα)= (tanα+1)/(1-cosα)=
=(8/15+1)/(1-cosα)= 23/(15*(1-cosα)) (1)
As known cos²α=1-sin²α (divide by cos²α both sides of equation)
cos²a/cos²α=1/cos²α-sin²α/cos²α
1=1/cos²α-tg²α
1/cos²α=1+tg²α
cos²α=1/(1+tg²α)
cosα=sqrt(1/(1+tg²α))= +-sqrt(1/(1+64/225))=+-sqrt(225/(225+64))=
=+-sqrt(225/289)=+-15/17 (2)
Substitute in (1) cosα by (2):
1st use cosα=15/17
1) 23/(15*(1-cosα)) =23/(15*(1-15/17))= 23*17/2=195.5
2-nd use cosα=-15/17
2)23/(15*(1-cosα)) =23/(15*(1+15/17))= 23*17/32=12 7/32
