Which expression is equivalent to 10a + 6? 10(a + 3) 2(5a + 3) 5(5a + 1) 10(a + 1)

2 answers:

6 0

Hey there!
We can factor 10a+6 to get our answer.
Remember factoring as the opposite of the distributive property
(i.e factored would by ab+ac and distributive would be a(b+c)).
We can't have <em>a</em> on the outside, since the second term in our expression doesn't contain an that variable.
We take our greatest common factor, or GCF, so it's factored completely.
We have:
2(5a+3).
Hopefully this helps

madreJ [45]2 years ago

5 0

Answer:

2(5a+3).

Step-by-step explanation:

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18. 0

Step-by-step explanation:

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7/12 divided by 2/5 I need help thanks and bye

Law Incorporation [45]

hey there!

7/12 divided by 2/5 is 7/30

first you have to find the reciprocal for 2 / 5 can do that by flipping 2 / 5 you'll then get 5 / 2 now the rest is simple all you have to do is multiply 7/12 by 2/5 so you multiply 7 by 2 and you get 14 that's your numerator 10 divide 12 by 5 and you'll get 60 so now you have 14 / 60 7 / 30 now you simplify and you get

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3 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 $x = 0$ .