When you see the fraction 33/16, you realize that 33 divided by 16 is 2 remainder 1, so you add the 2 to the original 6, then put the remainder 1 on top of 16 ,which gives you 8 1/16
Lim x->1 (x^2+8x-2)
substitute x =1
= 1^2 +8(1)-2
= 1 +8 -2
= 7
So the limit = 7
Please give brainliest
1 equals 3 over 4
2 equals 1 over 6
3 equals 1 over 5
4 equals 1 over 3
5 equals 5 over 11
6 equals 1 over 4
7 equals 1 over 5
8 equals 1 over 9
9 equals 2 over 5
10 equals 1 over 3
The denominator is 14-x.
Since the denominator cannot be equal to zero, you take the denominator, set it equal to zero, and solve.
This ill give you the value or values that must be eliminated (cannot be in the domain).
14 - x = 0
add x to both sides
14 = x
The domain is x ≠ 14
OR
(- ∞, 14) ∪ (14, ∞)
OR
All real numbers except 14
Answer:
firt thing i would do is take out things that are alike but if these are the options then i would a) multiply the first equation by -1
