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Rufina [12.5K]
3 years ago
5

Which of the following functions is graphed?

Mathematics
2 answers:
lora16 [44]3 years ago
6 0

Answer:

c

Step-by-step explanation:

SashulF [63]3 years ago
6 0
C. Y=lx-2l-3 is the answer
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Please help!
Hatshy [7]

Answer:

each side is 6.

Step-by-step explanation:

a hexagon has six sides and 36 divided by 6 is 6

3 0
3 years ago
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ASHA 777 [7]
Median, upper quartile, and maximum are all incorrect
5 0
3 years ago
100 POINTS PLEASE HELP
yanalaym [24]
I believe it is A but i am not 100% sure
5 0
3 years ago
ASAP!!! PLEASE HELP & EXPLAIN.
Vika [28.1K]

Answer:

B

Step-by-step explanation:

the domain of a graph consists of all the input values shown on the x-axis.

so because the only parts shown on the x axis are 6 and -6 that is the answer!

hope this helps!

4 0
2 years ago
Which function is undefined for x = 0? y=3√x-2 y=√x-2 y=3√x+2 y=√x=2
Andre45 [30]

For this case, we must indicate which of the given functions is not defined forx = 0

By definition, we know that:

f (x) = \sqrt {x} has a domain from 0 to infinity.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. For it to be defined, the term within the root must be positive.

Thus, we observe that:

y = \sqrt {x-2} is not defined, the term inside the root is negative when x = 0.

While y = \sqrt {x + 2} if it is defined for x = 0.

f(x)=\sqrt[3]{x}, your domain is given by all real numbers.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. In the same way, its domain will be given by the real numbers, independently of the sign of the term inside the root.

So, we have:

y = \sqrt [3] {x-2} with x = 0: y = \sqrt [3] {- 2} is defined.

y = \sqrt [3] {x + 2}with x = 0: y = \sqrt [3] {2}in the same way is defined.

Answer:

y = \sqrt {x-2}

Option b


6 0
3 years ago
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