Begin by finding the lowest point the quadratic equation can be, the vertex;
x²-1= is just a translation down of the graph x²
vertex; (0, -1) and since the graph of x² would extend to infinity beyond that point, we can say {x| x≥0} for domain and {y| y≥-1}.
For the linear equation, it is possible to have all x and y values, therefore range and domain belong to all real numbers.
Hope I helped :)
Answer:

Step-by-step explanation:
You can start by dividing both sides by 2, getting
.
Answer:
cos
(
5
π
12
)
=
√
2
−
√
3
2
Explanation:
By the half angle formula:
XXXX
cos
(
θ
2
)
=
±
√
1
+
cos
(
θ
)
2
If
θ
2
=
5
π
12
XXXX
then
θ
=
5
π
6
Note that
5
π
6
is a standard angle in quadrant 2 with a reference angle of
π
6
so
cos
(
5
π
6
)
=
−
cos
(
π
6
)
=
−
√
3
2
Therefore
XXXX
cos
(
5
π
12
)
=
±
⎷
1
−
√
3
2
2
XXXXXXXXXXX
=
±
⎷
2
−
√
3
2
2
XXXXXXXXXXX
=
±
√
2
−
√
3
4
XXXXXXXXXXX
=
±
√
2
−
√
3
2
Since
5
π
12
<
π
2
XXXX
5
π
12
is in quadrant 1
XXXX
→
cos
(
5
π
12
)
is positive
XXXX
XXXX
XXXX
(the negative solution is extraneous)
answer in pic more explain
Answer:
Opposite/Adjacent
Step-by-step explanation:
ok?
I'm gonna hazard a guess at 309.
121+121=242
242-224=18
224-121=103
103-18=85
224+85=309