Answer: No puedo responder a esto sin que me muestres el problema.
Answer:
y = 6
Step-by-step explanation:
when they work together, you square the 3 and get 6
Step-by-step explanation:
cj2ckbigckb1kdjlfd2ohsohxo2feih2xgpu
Answer:
x = 1 is not part of the domain
Step-by-step explanation:
The denominator of g(x) cannot be zero as this would make g(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be.
solve
x - 1 = 0 ⇒ x = 1 ← restricted value
The current Brainliest answer seems to be answering the question "Every integer is a multiple of which number?" rather than the question presented here.
We say that one number is a <em>multiple </em>of a second number if we can get to the first one by <em>counting by the second</em>. For example, 18 is a multiple of 6 because we can reach it by counting by 6's (6, 12, <em>18</em>). Note that, for any number we want to count by, we can always start our count at 0.
By 2's: 0, 2, 4, 6, 8
By 6's: 0, 6, 12, 18
By 7's: 0, 7, 14, 21
Because we can always "reach" 0 regardless of the integer we're counting by, we can say that <em>0 is a multiple of every integer</em>.
More formally, we say that some number n is a multiple of an integer x if we can find another integer y so that x · y = n. By this definition, 18 would be a multiple of 6 because 6 · 3 = 18, and 3 is an integer. We can use the property that the product of any number and 0 is 0 to say that x · 0 = 0, where x can be any integer we want. Since 0 is also an integer, this means that, by definition, 0 is a multiple of every integer.
