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| Date post: | 28-Feb-2018 |
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7/25/2019 Maths 2000
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1 3
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1 3
1
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19 22
20 23
20 22
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x
4 A
O
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x
y
1 2 3
1
2
3
4
5
x
y
1 2 3
1
2
3
4
5
Température
A
Pression
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1
2
1 2
x 2 1/2 x
^ *
x/2x^2+
-
1/4
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x
image 1
image 2
A M D
C
N
B
P
X
?
6
2
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x
y
0
1
2
3
4
5
6
7
8
9
0 0.5 1 1.5 2 2.5 3 3.5 4
x*(6-x)
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1
10
1 2 3 4 5 6 7 8 9 10
4
7
8
x
y
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f(x)=x+1
f(x)=x/2+1
f(x)=1
f(x)=-x/2+1
f(x)=-x+1
f(x)=-2x+1
0 1 2 x
y
Æ
5%
100 m
5 m
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0
8
8
8
8-x
x2
0
1
1 2 3
2
x
y
C2
C1
X2
4 5
AB
C
x
y
M
x
℄
1
1 2 3
2
x
y
-X2
X2
2
X2
22
X
x
y
1
1
1/X
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80x
8
8
0 01
8
x
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75-5
-70
x
5<7-7<-5
23 23 75
5<7
5- 7- 5+ 7+
75
5<7
2x52x70
-5-72x(-5)2x(-7)
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A
A
1
1
aire= 1xA
aire=AxA
aire = 1xA
aire = AxA
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0-3
5|5|
|-3|
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V
V
V
UU
U
U+V
U+V
Parallèlogramme
L’origine du 2e est
placé à l’extrémité
du 1er
DEUX FACONS DE TRACER LA SOMME
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A
B
C
M
A
B
C
M
N
Q
P
A B C
B A C
Æ
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Hypotheses On veut obtenir....
A(xa;ya)
B(xb;yb)
AB(xb-xa;yb-ya)
Traduction: OA=xa i + ya j
OB= xb i + yb j
AB = (xb-xa) i + (yb-ya) j
Démonstration
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U2
1V
-3
-1,5
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b=33-2j
substitution
b+2j=33
2b+3j=56
2(33-2j)+3j=56
66 -j=56
j=10
(1)
(2)
On remplace dans
(1) ou dans (2)
b+2x10=33 donc b=13
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Point ne verifiant ni (1) ni (2)
Point verifiant (1) mais pas (2)
Point verifiant (2) mais pas (1)
Solution du systeme (Point verifiant (1) ET (2))
b
j
℄
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A B
CD
I
DD’
I
A
B
C
D
D’
K
L
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DD’
A
BB’
C’
C
O
A
B
M
A B
C
AC
AB
D
AB+AC=AD
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Æ
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