Answer:
Step-by-step explanation:
Step 1: Add 4n to both sides.
n+4n=−4n−1+4n
5n=−1
Step 2: Divide both sides by 5.
5n
5
=
−1
5
n=
−1
5
Answer:
φ ≈ 1.19029 radians (≈ 68.2°)
Step-by-step explanation:
There are simple formulas for A and φ in this conversion, but it can be instructive to see how they are derived.
We want to compare ...
y(t) = Asin(ωt +φ)
to
y(t) = Psin(ωt) +Qcos(ωt)
Using trig identities to expand the first equation, we have ...
y(t) = Asin(ωt)cos(φ) +Acos(ωt)sin(φ)
Matching coefficients with the second equation, we have ...
P = Acos(φ)
Q = Asin(φ)
The ratio of these eliminates A and gives a relation for φ:
Q/P = sin(φ)/cos(φ)
Q/P = tan(φ)
φ = arctan(Q/P) . . . . taking quadrant into account
__
We can also use our equations for P and Q to find A:
P² +Q² = (Acos(φ))² +(Asin(φ))² = A²(cos(φ)² +sin(φ)²) = A²
A = √(P² +Q²)
_____
Here, we want φ.
φ = arctan(Q/P) = arctan(5/2)
φ ≈ 1.19029 . . . radians
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
}
The answer would probably be 16 meters
Answer:
I don't know if it's correct
but I tried
Step-by-step explanation:
let the angle be x
20+4x+x=180
20+5x=180
5x=160
x=32
