The domain of any graph depends on the vertex as well as the asymptotes. You can easily tell if the domain is all real numbers by looking at the graph and seeing that it goes clearly from -infinity to +infinity. If there is an asymptote that blocks it, then it will not be all real numbers.
Answer:

Step-by-step explanation:
we are given

we can simplify left side and make it equal to right side
we can use trig identity


now, we can plug values

now, we can simplify



now, we can factor it

![\frac{(sin(a)+cos(a))[3-4(sin^2(a)+cos^2(a)-sin(a)cos(a)]}{sin(a)+cos(a)}](https://tex.z-dn.net/?f=%5Cfrac%7B%28sin%28a%29%2Bcos%28a%29%29%5B3-4%28sin%5E2%28a%29%2Bcos%5E2%28a%29-sin%28a%29cos%28a%29%5D%7D%7Bsin%28a%29%2Bcos%28a%29%7D%20)
we can use trig identity

![\frac{(sin(a)+cos(a))[3-4(1-sin(a)cos(a)]}{sin(a)+cos(a)}](https://tex.z-dn.net/?f=%5Cfrac%7B%28sin%28a%29%2Bcos%28a%29%29%5B3-4%281-sin%28a%29cos%28a%29%5D%7D%7Bsin%28a%29%2Bcos%28a%29%7D%20)
we can cancel terms

now, we can simplify it further




now, we can use trig identity

we can replace it

so,

Answer:
Step-by-step explanation:
There is nothing in this diagram to suggest any particular values for x and y.
__
If we assume the figure is intended to be a kite, symmetrical about its horizontal diagonal, then each segment is congruent to the one above it.
x = 12
y = 20
The probability of randomly selecting a can of pink paint is P = 1/3, so the correct option is A.
<h3>What is the chance that he will paint his bedroom pink?</h3>
Assuming that all the cans of paint have the same probability of being randomly selected, the probability that he will choose a pink can is equal to the quotient between the number of pink cans and the total number of cans.
There are 2 cans of white, 4 cans of green, and 3 cans of pink, so a total of 9 cans.
Then the probability is:
P = 3/9 = 1/3.
The correct option is A.
If you want to learn more about probability:
brainly.com/question/251701
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Answer:
6.33... and 0.333...
Step-by-step explanation:
The quadratic formula is
.
It is important because while some quadratics are factorable and can be solved not all are. The formula will solve all quadratic equations and can also give both real and imaginary solutions. Using the formula will require less work than finding the factors if factorable. We will substitute a=9, b=-54 and c=-19.

We will now solve for the plus and the minus.
The plus,,,
and the minus...
