The answer to all of them is yes.
6) The lengths of AB and CD using the distance formula. (because congruent segments have equal length)
7) The slopes of AB and CD are equal using the slope formula. (because parallel segments have equal slopesL
8) The slopes of AB and CD are negative reciprocals using the slope formula. (because perpendicular lines have slopes that are negative reciprocals)
9) The two segments that CD is split into by AB have equal length using the distance formula. (because a segment bisector splits a segment into two congruent segments, and congruent segments have equal length)
10) Angles CAB and DAB have the same measure using the angle between two lines formula. (Because an angle bisector splits an angle into two congruent angles, and congruent angles have equal measure)
11) Angles A and B have the same measure using the angle between two lines formula. (Because an angle bisector splits an angle into two congruent angles, and congruent angles have equal measure)
12) The lines that form angle A have slopes that are negative reciprocals using the slope formula. (because perpendicular lines have slopes that are negative reciprocals, and perpendicular lines form right angles)
13) The lengths of AB and AC combined equal the length of AC using the distance formula.
14) Two sides of triangle ABC have equal length using the distance formula.
15) All four sides of ABCD have the same length using the distance formula.
16) Letting AB and CD meet at E, the distance formula says AE=BE and CE=DE.
Answer:
3x^2 + 10x - 8 = 0.
Step-by-step explanation:
In factor form it would be:
(x + 4)(3x - 2) = 0 (because x + 4 = 0 gives x = -4 and 3x - 2 = 0 gives x = 2/3);
Now change to standard form:
= 3x^2 - 2x + 12x - 8 =0
= 3x^2 + 10x - 8 = 0.
Step-by-step explanation:
Here we have
f
(
x
)
=
2
x
2
(
x
2
−
9
)
, which can be factorized as
f
(
x
)
=
2
x
2
(
x
+
3
)
(
x
−
3
)
As there is no common factor between numerator and denominator, there s no hole.
Further vertical asymptotes are
x
=
−
3
and
x
=
3
and as
f
(
x
)
=
2
x
2
(
x
2
−
9
)
=
2
1
−
9
x
2
, as
x
→
∞
,
f
(
x
)
→
2
, hence horizontal asymptote is
y
=
2
.
Observe that
f
(
−
x
)
=
f
(
x
)
and hence graph is symmetric w.r.t.
y
-axis. Further as
x
=
0
,
f
(
x
)
=
0
. Using calculas we can find that at
(
0
,
0
)
there is a local maxima as
d
y
d
x
=
−
36
x
(
x
2
−
9
)
2
and at
x
=
0
it is
0
. Further while for
x
<
−
3
and
x
>
3
, function is positive, for
−
3
<
x
<
3
function is negative.
Now take a few values of
x
say
{
−
10
,
−
7
,
−
4
,
−
2
,
−
1
,
1
,
2
,
4
,
7
,
10
}
and corresponding values of
f
(
x
)
are
{
2
18
91
,
2
9
20
,
4
4
7
,
−
1
3
5
,
−
1
4
,
−
1
4
,
−
1
3
5
,
4
4
7
,
2
9
20
,
2
18
91
}
I think the answer to this one would be B
The place value is the tenths.
