The value of B is not similar to the value of C.

Mathematics

1 answer:

levacccp [35]2 years ago

6 0

What type of question is this

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1 What is the value of the expression below?

lozanna [386]

Answer:

a - 1/50 is the ans

1/5 divided by 10 = 0.02

1/50 is also 0.02

7 0

3 years ago

Edge: Which equation has a focus at (–6, 12) and directrix of x = –12?

Volgvan

Answer:

C) (y - 12)^2 = 12(x+9)

8 0

2 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=3a-4%5Cleq5" id="TexFormula1" title="3a-4\leq5" alt="3a-4\leq5" align="absmiddle" class="latex

Gnesinka [82]

It depends, what does a mean? If you know just add what a means to 3 then subtract 4.

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

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:

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