Casio FX-890P 取扱説明書

fx-82MS

fx-83MS

fx-85MS

fx-270MS

fx-300MS

fx-350MS

J

P

http://www.casio.co.jp/edu/

AB2Mode

=

................ COMP

.................... Deg

............ Norm 1

............ ab/c

......... Dot

.....1

– 1 –

.....2

•

1

•

2

– 2 –

– 3 –

u

– 4 –

5

fx-82MS ....................................... 2 3

fx-83MS 85MS 270MS 300MS

350MS ....................................... 3 LR44

– 5 –

– 6 –

3 x x2x3x1x!

Ran#

π

nPr nCr

Pol x y Rec r

θ

Eng

– 7 –

– 8 –

COMP F 1

SD F 2

REG F 3

FDeg Rad

-

REG

SD

– 9 –

A B 2Mode =

COMP

Deg

Norm 1

ab/c

Dot

SD REG

COMP Deg Rad Gra

79

79 1 1

+ , - \ 1 1

A D 2

1 1

179

73

g

– 10 –

1cos60 sin60

W 60

e e e S

2369   2 369  2[

369 - - 2

e e [

32.362sin2.362

2 l 36 K

e e e e e A K S

AK

=

[

e r

e

r

0.

369××2–

0.

cos 60–

0.

sin 60

–

0.

369×2

0.

2.362–

0.

sin

2

.362

– 11 –

-4.123.586.4 = 21.1496

4.123.587.1 = 7.6496

4.12 - 3.58 + 6.4 =

e e e e e , 7.1

=

t

3 - 4 =

t

[ [

128

5

AB

r e

-14102.3 1402.3

14 \ 0 - 2.3 =

21.1496

4.12×3.58+6.

21.1496

12×3.58-7.1

–

7.6496

4.12×3.58-7.

12.

3×4

0.

–

7.6496

4.12×3.58-7.

aMthERROR

– 12 –

r e

e A K 1

=

-23 4

2 + 3 p \ g - 4 =

=

10

10 2

Norm 1

Norm 1 1010 10–2

Norm 2

Norm 2 1010 10–9

0.

0

14÷×2.3

0.

14÷1×2.3

0

3.22

14÷1×2.3

0

5.

2+3

Disp

20.

Ans ×4

– 13 –

Norm 1 Norm 2

F

1r1, 2

1Dot 3

2Comma 3

1

2

3

Fix Sci

N

or

m

5. –03

0.005

1

Disp

3 1 Norm 1 2Norm

2

1 \ 200 =Norm 1

Norm 2

Norm 1

F

– 14 –

COMP

AB3All =

COMP

COMP ............................................................ F 1

sin 2.34  105 → S 2.34 e D 5

1234.553  –

25.5

23 + 4.5 , 53 =

256(12)(2.5) 

268.8

56 - R D 12 T \ R D 2.5 T =

323(11020) 

6.666666667



10 19

2 \ 3 -1 e 20 =

47845 

36

7 - 8 , 4 - 5 =

5 =

0.3

6 \ R 4 - 5 T =

62[76(54)]

122

2 - R 7 + 6 -

R 5 + 4 T T =

= T

6

45

– 15 –

10

1  

2 C 3 + 1 C 5 =

23 1 

4

3 C 1 C 4 +

1 C 2 C 3 =

3 2 C 4 =

4 1.6 

2.1

1 C 2 + 1.6 =

↔

2

12.75  2 →

2.75 =

C

A B

13 15.

2

4

1

2

1

2

3

1

5

1

4

2

3

11

12

411 12.

13

15

11

4

3

4

11 4.

2 3 4.

2.75

– 16 –

2↔ 0.5 ↔

1 C 2 =

C

↔

-1 ↔

1 C 2 C 3 =

A B

F

1 2.

0.5

1 2.

1

2

1 2 3.

5 3.

1 2 3.

2

3

5

3

1

Disp

1 1, 2

1ab/c

2d/c

d/c

11500 12

(180)

1500 - 12 A v

2660 880

(75%)

660 \ 880 A v

32500 15

(2875)

2500 - 15 A v +

– 17 –

43500 25

(2625)

3500 - 25 A v ,

5168 98 734 20

(800)

168 + 98 + 734 = g A J 1

p 1 - 20 A v ,

, v

6500g 300g

(160%)

300 + 500 A v

740 46

48

(15%, 20%)

46 , 40 A v

e e e e e e 8 =

60 10

12.258 60 ↔ 10

2.258 =

2.258

A O

2°15°28.8

I

2.258

212° 34’ 56” × 3.45

12 I 34 I 56 I - 3.45 =

43°24°31.2

– 18 –

F

1 3

1Fix

2Sci

3Norm

1200  7  14  400

200 \ 7 - 14 =

F..... 1Fix 3

200 \ 7 =

- 14 =

200 \ 7 =

A Q

- 14 =

F

.....

3Norm

1

21  3 2Sci 2

F..... 2Sci 2 1 \ 3 =

F.....3Norm 1

1

2

3

Fix Sci

N

or

m

400.

400.000

28.571

399.994

400.00000

FIX

3.3–01

SCI

– 19 –

COMP

COMP ............................................................. F 1

=

g

-123456579 123 + 456 =

789579210

789 , g =

12 2

= A v|A { A j ( = A~F, M, X,

Y)

-3412 3.14

3 - 4 =

\ 3.14 =

x2x3x–1 x! DRG' 

^(xy)x  nPr nCr

210.

789–

A

ns

12.

3×4

3.821656051

A

ns÷

3

.14

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