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Antwoorden Wiskunde B (Moderne wiskunde 11 editie) 4 VWO Hoofdstuk 4 $3.33   Add to cart
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Antwoorden Wiskunde B (Moderne wiskunde 11 editie) 4 VWO Hoofdstuk 4

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  • Course
  • Wiskunde B
  • Level
  • VWO / Gymnasium

Antwoorden Moderne Wiskunde 11e editie Wiskunde B VWO Hoofdstuk 4

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Document information

  • March 13, 2017
  • 14
  • 2015/2016
  • Answers
  • Unknown

Subjects

  • moderne wiskunde wiskunde b vwo

Written for

  • Secondary school
  • VWO / Gymnasium
  • Wiskunde B
  • 4
All documents for this subject (269)

2  reviews

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By: marjannevandiermen • 4 year ago

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By: mariowezeman • 6 year ago

Translated by Google

May be a bit out of knitted

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Content preview

13 maart 2017




Hoofdstuk 4: Exponentiële functies
V-1.
a.  1,3
156
120
203
156
 1,30 263
203
 1,30 342
263
 1,30
De groeifactor per uur is 1,30
b. De groeifactor per dag is 1,3024  542,8
c. op tijdstip t  1: 120
1,3
 92 en op tijdstip t  2 : 1,3
92
 71
d. 120  1,3t  660
Voer in: y1  120  1,3 x en y 2  660 intersect: x  6,5
Na bijna 6,5 uur waren er 660 bacteriën.

V-2.
a. gweek  1,437  12,23
1
b. ghalve dag  1,43 2  1,20
1
N(t )  15310  1,4324  15310  1,015t
t
c.
4 34
d. N(4 34 )  15310  1,015  16433

V-3.
a. N (t )  23  1,01324t  23  1,36t
1
N(t )  254  13,7365  254  1,007t
t
b.
c. N(t )  3750
1,95
 1,9t  151 1,9t

V-4.
1
a. Na een half jaar heb je de beginhoeveelheid vermenigvuldigd met 2 2  2 .
1
b. gmaand  212
c. 2 3  1
8

d. Dan moet je delen door 212  4096
1 31
e. 8 3 2  23  2 3  2 3 dus t  3 31

V-5.
121 5 21
( 213 ) 2  24 (23 ) 2  24  2
1 1
a. 1
16
1
8
 1
24
2
b. 24  213  24  23  21
1 1 4 2
(24 )7 (22 )3 4 2 5
16 3 4
   2 7  2 3  25  2
7
2 7 23 6 21
c. 1 1
25
32 25


43  8 2  2  (22 )3  (23 ) 2  22  26  2
41 41 1 13 21 1
d.  22  28

V-6.
 31
g 2 
1
a. g 3  12 b. g4  2 c. 1
4 d. 15g 3
1  21 1  31
g  12 3
g  24  16 g  ( 41 ) 4 2
2
g  51
g  ( 51 )3  125



1
Uitwerkingen 4 vwo wiskunde B, hoofdstuk 4

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