Hs13 Algebraïsche vaardigheden

Diagnostische Toets Wis C VWO Hoofdstuk 13 Algebraïsche vaardigheden Getal en Ruimte wiskunde uitwerkingen www.uitwerkingensite.nl
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13.8 Diagnostische toets
bladzijde 50
1
K = aq + b met a =
K = 1,75q + b
q = 200 en K = 575
∆K 1030 − 575
=
= 1,75
∆q
460 − 200
⋅ 200 + b = 575
} 1,75
350 + b = 575
Dus K = 1,75q + 225.
2
b = 225
a Los op 2,12t + 253 = 1,86t + 274
0,26t = 21
t ≈ 80,8
K
KII
KI
t
O
Uit de schets volgt dat bij 81 uur of meer zonnebank II goedkoper is dan zonnebank I.
en KII − KI = 10.
b Los op KI − KII = 10
2,12t + 253 − (1,86t + 274) = 10
1,86t + 274 − (2,12t + 253) = 10
2,12t + 253 − 1,86 − 274 = 10
1,86t + 274 − 2,12t − 253 = 10
0,26t − 21 = 10
− 0,26t + 21 = 10
0,26t = 31
− 0,26t = −11
t ≈ 119,2
t ≈ 42,3
Verschil is minder dan 10 euro bij gebruik tussen 42 en 119 uur per jaar.
3
a a + 2b = 36
2b = −a + 36
b = − 0,5a + 18
P = 20 − 0,3a + 0,8b
b P = 7ab − 2a
b = 3a − 4
0,3a + 0,8(−0,5a + 18)
} PP == 2020 −− 0,3a
− 0,4a + 14,4
P = 34,4 − 0,7a
− 4) − 2a
} PP == 7a(3a
21a − 28a − 2a
2
P = 21a2 − 30a
4
a
1
x
2
x
1
=
1
2
=
1
1
22
=x
1
−2 2
x ⋅x
x
12
3,8
b (x ) ⋅ x = x ⋅ x = x15,8
c (x2a − 1)4 ⋅ x3 = x8a − 4 ⋅ x3 = x8a − 1
2,4 5
2
3,8
1
3
1
3
1
5
a y = 2x3 ⋅ x = 2x 3 ⋅ x 2 = 2x 2 , dus y = 2x 2 .
1
11
11
22
−2
−2
−2
−3
b y = 33 ⋅⋅ xx == 2x
⋅⋅ xx22 == 2x
2 x−3
2 x 22,,dus y = 2x 2
xx
c y = (3x−1,2)3 ⋅ 2x4,6 = 27x−3,6 ⋅ 2x4,6 = 54x, dus y = 54x.
6
a y = 5 ⋅ x3,2 geeft 5 ⋅ x3,2 = y
x3,2 = 15 y
1
x = ( 15 y ) 3,2
1
1
x = ( 15 ) 3,2 ⋅ y 3,2
x = 0,60 ⋅ y0,31
b y = 5 ⋅ 3 x 2 geeft 5 ⋅ 3 x 2 = y
2
x 3 = 15 y
1
5
x = ( y)
3
2
c y = 5 ⋅ (3x)2,1
1
( 2
)
3
3
x = ( 15 ) ⋅ y 2
x = 0,09 ⋅ y1,5
geeft 5 ⋅ (3x)2,1 = y
(3x)2,1 = 15 y
1
3x = ( 15 y ) 2 ,1
1
1
x = 13 ⋅ ( 51 ) 2 ,1 ⋅ y 2 ,1
x = 0,15 ⋅ y0,48
7
a N = 15 ⋅ 2,63t − 1
N = 15 ⋅ 2,63t ⋅ 2,6−1
N = 15 ⋅ 2,6−1 ⋅ (2,63)t
N = 5,77 ⋅ 17,58t
b N = 600 ⋅ 0,84t − 3
N = 600 ⋅ 0,84t ⋅ 0,8−3
N = 600 ⋅ 0,8−3 ⋅ (0,84)t
N = 1172 ⋅ 0,41t
c N = 500 ⋅ 0,20,05t − 2
N = 500 ⋅ 0,20,05t ⋅ 0,2−2
N = 500 ⋅ 0,2−2 ⋅ (0,20,05)t
N = 12 500 ⋅ 0,92t
bladzijde 51
8
a gdag = 1,36, dus gweek = 1,367 ≈ 8,61.
De toename per week is 761%.
1
b gdag = 1,36, dus guur = 1, 36 24 ≈ 1,013.
De toename per uur is 1,3%.
9
1
a g10 jaar = 0,75, dus gjaar = 0, 7510 ≈ 0,972.
De afname per jaar is 2,8%.
b g10 jaar = 0,75, dus g25 jaar = 0,752,5 ≈ 0,487.
De afname per 25 jaar is 51,3%.
10 a N1 = at + b met a =
∆N 3900 − 3000
=
= 300.
∆t
8−5
N1 = 300t + b
t = 5 en N1 = 3000
} 300 ⋅ 5 + b = 3000
N2 = b ⋅ 1,0914t
t = 5 en N2 = 3000
} b ⋅ 1,0914 = 3000
b = 1500
Dus N1 = 300t + 1500.
3900
b N2 = b ⋅ gt met g3 tijdseenheden =
= 1,3
3000
1
3
gtijdseenheid = 1, 3 ≈ 1,0914
5
b=
3000
≈ 1940
1, 09145
Dus N2 = 1940 ⋅ 1,0914t.
c N1 = 9000 geeft 300t + 1500 = 9000
300t = 7500
t = 25
Op t = 25 is N2 = 1940 ⋅ 1,091425 ≈ 17 270.
11
a 3 + 2 ⋅ log(x) = 7
2 ⋅ log(x) = 4
log(x) = 2
x = 102 = 100
b log(2x − 300) = 4
2x − 300 = 104
2x = 10 300
x = 5150
c 5 + 3 ⋅ log(x) = 2
3 ⋅ log(x) = −3
log(x) = −1
x = 10−1 = 0,1
12 a f (6) = 2 + log(9) ≈ 2,95
b f (x) = 5 geeft 2 + log(x + 3) = 5
log(x + 3) = 3
x + 3 = 103
x = 997
c Voer in y1 = 2 + log(x + 3) en y2 = − 0,5x + 1.
De optie intersect geeft x = −2.
y
ƒ
g
x
O
–2
x = –3
f (x) < g(x) geeft −3 < x < −2.
1
1
2
13 a 5 ⋅ 3log(4) −
⋅ 3log(16) = 3log(45) − 3 log(16 2 ) = 3log(1024) − 3 log( 16) =
 1024 3
log(1024) − 3log(4) = 3 log 
= log(256)
 4 
3
b log(x + 145) = 1 + log(x + 10)
log(x + 145) = log(10) + log(x + 10)
log(x + 145) = log(10(x + 10))
log(x + 145) = log(10x + 100)
x + 145 = 10x + 100
−9x = − 45
x=5
c 5 ⋅ log(N) = 10 − 4P
log(N) = 2 −
N = 10
2−
4
5
P
4
P
5
N = 102 ⋅ 10
4
−5P
−
4
N = 102 ⋅ (10 5 )
P
Je krijgt N = 100 ⋅ 0,158P.
d F = 560 ⋅ 1,175t
log(F) = log(560 ⋅ 1,175t)
log(F) = log(560) + log(1,175t)
log(F) = log(560) + t ⋅ log(1,175)
Je krijgt log(F) = 0,07t + 2,75.
14 a Buurgemeente A
Rechte lijn op logaritmisch papier, dus NA = b ⋅ gt.
70000
t = 0 en N = 20 000
1
g50 jaar =
= 32
t = 50 en N = 70 000
20000
}
1
gjaar = (3 12 ) 50 ≈ 1,025
De formule is NA = 20 000 ⋅ 1,025t.
Buurgemeente B
Rechte lijn op logaritmisch papier, dus NB = b ⋅ gt.
20000
t = 0 en N = 200 000
g50 jaar =
= 0,1
t = 50 en N = 20 000
200000
}
1
gjaar = 0,150 ≈ 0,955
De formule is NB = 200 000 ⋅ 0,955t.
b Voer in y1 = 20 000 ⋅ 1,025x + 200 000 ⋅ 0,955x.
De optie minimum geeft x ≈ 41,4 en y ≈ 85 318.
In 1991 is het aantal minimaal. Er zijn dan 85 318 inwoners.

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