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Tasya [4]
2 years ago
9

2+2=4 :))))) Jk it’s fish

Mathematics
2 answers:
Anna [14]2 years ago
8 0

Answer:

Caca :0

Step-by-step explanation:

Goryan [66]2 years ago
4 0

Step-by-step explanation:

wow

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Profits were 18,447 last month and 14,300 the previous month. The rate of increase was what?
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a coin must have diameter of 27.13 millimeters.To the nearest hundredth, what is the circumference of this coin millimeters
Bas_tet [7]
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Which function is undefined for x = 0? y=3√x-2 y=√x-2 y=3√x+2 y=√x=2
Mkey [24]

For this case, we have to:

By definition, we know:

The domain of f (x) = \sqrt [3] {x} 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. Thus, it will always be defined.

So, we have:

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

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

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

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 whenx = 0.

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

Answer:

y = \sqrt {x-2}

Option b

6 0
3 years ago
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