Casio FX-5800P Supplement [es]
fx-5800P
Tillägg
Supplement
Suplemento
Ergänzung
Supplemento
J E S G I Sw KCh C
k
http://edu.casio.jp/
http://world.casio.com/edu/
付録
RJA516833-001V01
1-1
m
p
1.67262171×10
–27
kg
3-5
1-2
mn1.67492728×10
–27
kg
3-6
F
96485.3383 C mol
–1
1-3
me9.1093826×10
–31
kg
3-7
e
1.60217653×10
–19
C
1-4
m
1.8835314×10
–28
kg
3-8
N
A
6.0221415×10
23
mol
–1
1-5
a
0
0.5291772108×10
–10
m
4-1
k
1.3806505×10
–23
J K
–1
1-6
h
6.6260693×10
–34
J s
4-2
V
m
22.413996×10
–3
m3 mol
–1
1-7
5.05078343×10
–27
J T
–1
4-3
R
8.314472 J mol
–1
K
–1
1-8
927.400949×10
–26
J T
–1
4-4
C
0
299792458 m s
–1
2-1
H
1.05457168×10
–34
J s
4-5
C
1
3.74177138×10
–16
W m
2
2-2
α 7.297352568×10
–3
4-6
C
2
1.4387752×10
–2
m K
2-3
re2.817940325×10
–15
m
4-7
σ
5.670400×10–8 W m–2 K
–4
2-4
λc2.426310238×10
–12
m
4-8
ε
0
8.854187817×10
–12
F m
–1
2-5
γ
p
2.67522205×108 s
–1
T
–1
5-1
12.566370614×10
–7
N A
–2
2-6
λcp1.3214098555×10
–15
m
5-2
2.06783372×10
–15
Wb
2-7
λcn1.3195909067×10
–15
m
5-3
g
9.80665 m s
–2
2-8
R∞
10973731.568525 m
–1
5-4
G07.748091733×10
–5
S
3-1
u
1.66053886×10
–27
kg
5-5
Z
0
376.730313461 Ω
3-2
1.41060671×10
–26
J T
–1
5-6
t
273.15 K
3-3
–928.476412×10
–26
J T
–1
5-7
G
6.6742×10
–11
m3 kg–1 s
–2
3-4
–0.96623645×10
–26
J T
–1
5-8
atm
101325 Pa
–4.49044799×10
–26
J T
–1
N
µ
B
µ
µ
p
µ
e
µ
n
µ
0
µ
µ
µ
0
φ
#01
– 1 –
#02
a
=
n
Σy
i
–
a.Σx
i
b =
n.Σx
i
2
–
(
Σx
i
)
2
n.Σxiy
i
–
Σx
i
.
Σy
i
r
=
{
n.Σx
i
2
–
(
Σx
i
)
2
}{
n.Σy
i
2
–
(
Σy
i
)
2
}
n.Σxiy
i
–
Σx
i
.
Σy
i
m
y – b
a
=
n = ax + b
#03
n
Σy
i
c
= –
a
(
)
–
b
(
)
n
Σ
x
i
n
Σx
i
2
b
=
Sxx.Sx2x
2
– (Sxx
2)2
Sxy.Sx2x
2
– Sx
2
y.Sxx
2
a
=
Sxx.Sx2x
2
– (Sxx
2)2
Sx2y.Sxx – Sxy.Sxx
2
(
Σ
x
i
)
2
Sxx = Σx
i
2
–
n
Sxy = Σxiy
i
–
n
(Σx
i
.
Σy
i
)
Sxx
2
=
Σ
x
i
3
–
n
(
Σ
x
i
.
Σ
x
i
2
)
Sx2x
2
= Σx
i
4
–
n
(Σx
i
2)2
Sx2y = Σx
i
2
y
i
–
n
(Σx
i
2
.
Σy
i
)
m1
=
2a
– b +
b
2
– 4a
(
c
– y
)
m2 =
2a
–
b
–
b
2
– 4a
(
c
– y
)
n = ax2 + bx + c
#04
a =
n
Σy
i
– b
.
Σlnx
i
b =
n
.
Σ
(
lnx
i
)
2
–
(
Σlnx
i
)
2
n
.
Σ
(
lnx
i
)
y
i
– Σlnx
i
.
Σy
i
r =
{n
.
Σ
(
lnx
i
)
2
–
(
Σlnx
i
)
2
}{n
.
Σy
i
2
–
(
Σy
i
)
2
}
n
.
Σ
(
lnx
i
)
y
i
– Σlnx
i
.
Σy
i
m = e
y – a
b
n = a + blnx
– 2 –
a
= exp
(
)
n
Σlnyi – lnb.Σx
i
b = exp
(
)
n.Σx
i
2
–
(
Σx
i
)
2
n.Σx
i
lny
i
– Σx
i
.
Σlny
i
r =
{n
.
Σx
i
2
–
(
Σx
i
)
2
}{n
.
Σ
(
lny
i
)
2
–
(
Σlny
i
)
2
}
n.Σx
i
lny
i
– Σx
i
.
Σlny
i
m =
lnb
lny – lna
n = ab
x
a = exp
(
)
n
Σlnyi – b.Σlnx
i
b =
n.Σ
(
lnx
i
)
2
–
(
Σlnx
i
)
2
n.Σlnx
i
lny
i
– Σlnx
i
.
Σlny
i
r =
{n
.
Σ
(
lnx
i
)
2
–
(
Σlnx
i
)
2
}{n
.
Σ
(
lny
i
)
2
–
(
Σlny
i
)
2
}
n.Σlnx
i
lny
i
– Σlnx
i
.
Σlny
i
m = e
b
ln y – ln a
n = ax
b
#05
#06
#07
a = exp
(
)
n
Σlny
i
– b
.
Σx
i
b =
n.Σx
i
2
–
(
Σx
i
)
2
n.Σx
i
lny
i
– Σx
i
.
Σlny
i
r =
{n
.
Σx
i
2
– (Σx
i
)
2
}{n
.
Σ
(
lny
i
)
2
– (Σlny
i
)
2
}
n.Σx
i
lny
i
– Σx
i
.
Σlny
i
m =
b
lny – lna
n = ae
bx
– 3 –
b =
Sxx
Sxy
r =
Sxx.Syy
Sxy
S
xx = Σ (x
i
–1
)
2
–
Syy = Σy
i
2
–
S
xy = Σ(x
i
–1
)y
i
–
n
(Σx
i
–1
)
2
n
Σx
i
–1
.
Σy
i
n
(
Σy
i
)
2
a =
n
Σy
i
– b
.
Σx
i
–1
#08
m
=
y – a
b
n = a +
x
b
1
2
3
4
5
tan
=
θ
m
2 – m1
1 +
m1m2
(m
1m2 G
1)
a
=
t
2 – t1
(t2 >
t1 >
0)
a
b
c
A
B
C
S
=
n{2a + (n – 1)d}
2
#09
θ
y
x
y = m2 x + k
2
y = m
1
x + k
1
cos A =
2
bc
b
2
+ c2 – a
2
cos B =
2
ca
c
2
+ a2 – b
2
cos C =
2
ab
a
2
+ b2 – c
2
2 – 1
υ υ
S
= 0
t + at
2
1
2
(t
> 0)
υ
– 4 –
6
7
8
9
[(xp, yp)→(Xp, Yp)]
Xp = (x
p–x0
)cos + (yp–y0)sin
Yp
= (
y
p–y0
)cos – (xp–x0)sin
α
α
10
11
12
=
(M, T > 0)
3RT
M
[ +
+
Z
=
Const.]
P
γ
2g
P
2
= P
1
+
γ
(
2g
+ Z1 – Z
2
)
(υ, P, , Z > 0)
γ
[ +
+
Z
=
Const.
]
P
γ
2
2g
2g(P
1
– P2)
γ
[ +
+
Z
=
Const.]
P
γ
P
x = nCx
P
x
( 1 – P)
n–x
13
η
=
Q
1
– Q
2
Q
1
( Q
1
G 0)
X
P
= Rcos
+
X
A
α
Y
P
=
Rsin
+
Y
A
α
α
α
y
x
Y
X
(0, 0)
(x0, y0)
α
(x
p
, y
p
)
(X
p
, Y
p
)
(XA, YA)
(Xp, Yp)
(XB, YB)
α
R
υ
2
υ
1
2
– 2
2
υ
υ
υ
+
1
2
+
2g( Z1 – Z2)
υ
( , P, , Z > 0)
γ
υ
2
=
υ
2
2g
υ
γ
( , P, , Z > 0)
γ
υ
Z
2
= + + Z
1
P
1 – P2
1
2
–
2
2
2g
υυ
(
)
0 < P < 1
x = 0, 1, 2······
Pol(X
B
– XA, YB – YA)
– 5 –
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