We keep the 2x^2 because we can only subtract something from that if it also is squared. From there, we subtract the 7x from 2x^2-11. Since there is no previous x's in the 2x^2-11, we just make it -7x. So, without even continuing the problem, we see that A) is the correct answer because it is the only one with -7x.
So,
#1:

Convert to like improper fractions.

Add.

So, one solution could be

.
Another solution could by 9. There is also 10, 11, 12, etc., and all numbers in between.
#2:

Convert into improper fraction form.

Multiply.

Cross-cancel, and we have our final result.

k < 96
96 is not a solution.
95 is a solution.
So is 94, 93, 92, etc, and all numbers in between.
Answer:
i don't know
Step-by-step explanation:
sorry
The <u>correct answer</u> is:
As x→-∞, y→-3.
As x→∞, y→∞.
Explanation:
As our values of x get further into the negative numbers, the value of 2ˣ will approach 0. This is because raising a number to a negative exponent "flips" the number below the denominator and raises it to a power; we end up with smaller and smaller fractions, eventually so small that they nearly equal 0.
This will make the value of the function 0-3=-3.
As x gets larger and larger (towards ∞), the value of y, 2ˣ, continues to grow as well. Since it continues to grow exponentially, we say the value approaches ∞.
Yes they are proportionate