Answer:
2 vertical asymptotes occurring at x = 5 and x = -1
Step-by-step explanation:
given
recall that asymptotic occur at the locations that will make the equation undefined. In this case, the asymptote will occur at x-locations which will cause the denominator to become zero (and hence undefined)
Equating the denominator to zero,
(x-5)(x+1) = 0
(x-5) =0
x = 5 (first asymptote)
or (x+1) = 0
x = -1 (2nd asymptote)
Choices A, C, D have a y-value that is 5 more than the x-value, so those points are on the line defined by the first equation.
Choice D satisfies the requirement
... y = 2x - 13
... 23 = 2·18 - 13
... 23 = 36 - 13
The appropriate choice is ...
... D (18, 23)
<span>y=x-4
y=-x+6
---------------------
Substitute x - 4 for y in </span>y=-x+6
x-4=-x+6<span>
--------------------------------
Add X on each side
</span><span>x-4+x</span>=-<span>x+6+x
</span>2x - 4 = 6
--------------------------------------
Add 4 on each side
<span>2x-4+4</span>=<span>6+4
</span>2x = 10
------------------------------------------------
Divide by 2 on each side
2x ÷ 2 = 10 ÷ 2
x = 5
Now we have X
================================================================
To find Y we substitute 5 for x in y=<span>x-<span>4
</span></span>y = 5 - 4
y = 1
-----------------------------------------------------
Your answers are Y = 1 and X = 5
Answer:
Since the only numbers that are not 0 appear in the groups 5 < d ≤ 10 to 30 < d ≤ 35 the range is 5 < d ≤ 35.
