In the first equation, you can tell that the x's are being multiplied by 3 to get to the y's so the blank will be 9. The second table would not be a function since all the x values are not being multiplied by the same number to get the y value. Hope this helps!
If I am reading this right, it looks like the 10, 3, 2, 1 are Adjustments and the Adjusted TB should equal the difference. Make sure you know how to add and subtract the debit and credit adjustments correctly.
TB +/- Adj = ATB
Answer: do the math, 39x
Step-by-step explanation:2+4=6 times 7=42-19=23+7=30+9=39x
Answer:
The answer is 15
Step-by-step explanation:
I know and is not stupid
Answer:
Step-by-step explanation:
x
2
+
x
−
6
=
(
x
+
3
)
(
x
−
2
)
x
2
−
3
x
−
4
=
(
x
−
4
)
(
x
+
1
)
Each of the linear factors occurs precisely once, so the sign of the given rational expression will change at each of the points where one of the linear factors is zero. That is at:
x
=
−
3
,
−
1
,
2
,
4
Note that when
x
is large, the
x
2
terms will dominate the values of the numerator and denominator, making both positive.
Hence the sign of the value of the rational expression in each of the intervals
(
−
∞
,
−
3
)
,
(
−
3
,
−
1
)
,
(
−
1
,
2
)
,
(
2
,
4
)
and
(
4
,
∞
)
follows the pattern
+
−
+
−
+
. Hence the intervals
(
−
3
,
−
1
)
and
(
2
,
4
)
are both part of the solution set.
When
x
=
−
1
or
x
=
4
, the denominator is zero so the rational expression is undefined. Since the numerator is non-zero at those values, the function will have vertical asymptotes at those points (and not satisfy the inequality).
When
x
=
−
3
or
x
=
2
, the numerator is zero and the denominator is non-zero. So the function will be zero and satisfy the inequality at those points.
Hence the solution is:
x
∈
[
−
3
,
−
1
)
∪
[
2
,
4
)
graph{(x^2+x-6)/(x^2-3x-4) [-10, 10, -5, 5]}
