Answer:
The range is all real number y<3.
Step-by-step explanation:
So we have to have v(x) exist for input x.
Let's think about that. v(x)=1/x so the domain is all real numbers except 0 since you cannot divide by 0. v(x)=1/x will also never output 0 because the numerator of 1/x is never 0. So the range of v(x)=1/x is also all real numbers except y=0.
Now let's plug v into u:
The domain of will still have the restrictions of v; let's see if we see any others here.
Nope, there are no, others, the only thing that is bothering this function is still the division by x (which means we can't plug in 0).
Let's thing about what are y's value will not ever get to be.
Let's start with that fraction. -2/x^2 will never be 0 because -2 will never be 0.
So we will never have y=0+3 which means y will never be 3.
There is one more thing to notice -2/x^2 will never be positive because x^2 is always positive and as we know a negative divided by a positive is negative.
So we have (a always negative number) + 3 this means the range will only go as high as 3 without including 3.
The range is all real number y<3.
I would say it is 2000 but I'm not to sure
Answer:
Its B
Step-by-step explanation:
Answer:
Step-by-step explanation:
<u>Multiples of 16:</u>
<u>Multiples of 20:</u>
<u>Here are the options verified:</u>
- A. 4 and 20. Incorrect. It takes a while to get to 80 by 4's.
- B. 8 and 10. Incorrect. Same thing goes with A, but with 8's.
- C. 16 and 20. Correct.
- D. 12 and 40. Incorrect. 12 is not a multiple of 80.
Option C is your answer.