One way would be to notice that both functions have the same derivative and then find out what the "constant" is by plugging in x=0
x
=
0
.
Here's another way. Look at
π2−2arctanx−−√=2(π4−arctanx−−√)=2(arctan1−arctanx−−√).
π
2
−
2
arctan
x
=
2
(
π
4
−
arctan
x
)
=
2
(
arctan
1
−
arctan
x
)
.
Now remember the identity for the difference of two arctangents:
arctanu−arctanv=arctanu−v1+uv.
arctan
u
−
arctan
v
=
arctan
u
−
v
1
+
u
v
.
(This follows from the usual identity for the tangent of a sum.) The left side above becomes
2arctan1−x−−√1+x−−√.
2
arctan
1
−
x
1
+
x
.
The double-angle formula for the sine says sin(2u)=2sinucosu
sin
(
2
u
)
=
2
sin
u
cos
u
. Apply that:
sin(2arctan1−x−−√1+x−−√)=2sin(arctan1−x−−√1+x−−√)cos(arctan1−x−−√1+x−−√)
sin
(
2
arctan
1
−
x
1
+
x
)
=
2
sin
(
arctan
1
−
x
1
+
x
)
cos
(
arctan
1
−
x
1
+
x
)
Now remember that sin(arctanu)=u1+u2−−−−−√
sin
(
arctan
u
)
=
u
1
+
u
2
and cos(arctanu)=11+u2−−−−−√
cos
(
arctan
u
)
=
1
1
+
u
2
Then use algebra:
2⋅(1−x√1+x√)1+(1−x√1+x√)2−−−−−−−−−−√⋅11+(1−x√1+x√)2−−−−−−−−−−√=1−x1+x.
Answer: D
Step-by-step explanation:
Answer:
A. 0
Step-by-step explanation:
3ab+5b-6
= 3(-1)(3)+5(3)-6
=3(-3)+15-6
=-9+9
=0 ( you have to subtract because they don't have the same signs)
(a) Answer: One solution.
Explanation: two lines that have same slope with opposite sign will necessarily cross each other at a single point, as one is "pointing uphill" and the other "downhill."
(b) Answer: Infinitely many solutions.
Explanation:
2x + 3y = 5.5
4x + 6y = 11 | divide by 2
-->
2x + 3y = 5.5
2x + 3y = 5.5
--> equations are identical.
2x + 3y = 2x + 3y
so any (x,y) will satisfy this equation. This means infinitely many solutions.
(c) Answer: One solution
Explanation:
Continuing the two lines it becomes obvious they will cross at one point (the solution)
(d) Answer: No solution
Explanation: If the two lines are parallel, they will never cross (by definition of "parallel"). Therefore there will be no solution to the corresponding linear system.
