Vraag
Op hoeveel manieren kan men 22 jongens verdelen in 2 voetbalploegen?
Uitwerking
Kies 11 jongens uit 22 voor team A. Je hebt dan nog 11 jongens over
voor team B. Daarna nog delen door 2 omdat team A en B onderling
uitwisselbaar zijn.
\(\eqalign{\frac{{\left( {\begin{array}{*{20}c} {22} \\ {11} \\ \end{array}} \right)}}{2} = {\rm{352}}{\rm{.716}}}\)
Toelichting
...
Begrip en inzicht
...
zaterdag 20 juni 2020
donderdag 11 juni 2020
woensdag 10 juni 2020
HAVO wiskunde B - 2011 (pilot) 2e tijdvak
\(
\eqalign{
& 19. \cr
& (x - 7)^2 + (y - 9)^2 = 100 \cr
& \downarrow y = - {3 \over 4}x + 17{3 \over 4} \cr
& (x - 7)^2 + \left( { - {3 \over 4}x + 17{3 \over 4} - 9} \right)^2 = 100 \cr
& (x - 7)^2 + \left( { - {3 \over 4}x + 8{3 \over 4}} \right)^2 = 100 \cr
& (x - 7)^2 + {{\left( { - 3x + 35} \right)^2 } \over {4^2 }} = 100 \cr
& 16(x - 7)^2 + \left( { - 3x + 35} \right)^2 = 1600 \cr
& 16x^2 - 224x + 784 + 9x^2 - 210x + 1225 = 1600 \cr
& 25x^2 - 434x + 409 = 0 \cr
& \downarrow GR \cr
& x = 1 \vee x = 16{9 \over {25}} \cr
& \downarrow GR \cr
& \left\{ \matrix{
x = 16{9 \over {25}} \hfill \cr
y = - {3 \over 4} \cdot 16{9 \over {25}} + 17{3 \over 4} = 5{{12} \over {25}} \hfill \cr} \right. \cr
& Q\left( {16{9 \over {25}},5{{12} \over {25}}} \right) \cr}
\)
\eqalign{
& 19. \cr
& (x - 7)^2 + (y - 9)^2 = 100 \cr
& \downarrow y = - {3 \over 4}x + 17{3 \over 4} \cr
& (x - 7)^2 + \left( { - {3 \over 4}x + 17{3 \over 4} - 9} \right)^2 = 100 \cr
& (x - 7)^2 + \left( { - {3 \over 4}x + 8{3 \over 4}} \right)^2 = 100 \cr
& (x - 7)^2 + {{\left( { - 3x + 35} \right)^2 } \over {4^2 }} = 100 \cr
& 16(x - 7)^2 + \left( { - 3x + 35} \right)^2 = 1600 \cr
& 16x^2 - 224x + 784 + 9x^2 - 210x + 1225 = 1600 \cr
& 25x^2 - 434x + 409 = 0 \cr
& \downarrow GR \cr
& x = 1 \vee x = 16{9 \over {25}} \cr
& \downarrow GR \cr
& \left\{ \matrix{
x = 16{9 \over {25}} \hfill \cr
y = - {3 \over 4} \cdot 16{9 \over {25}} + 17{3 \over 4} = 5{{12} \over {25}} \hfill \cr} \right. \cr
& Q\left( {16{9 \over {25}},5{{12} \over {25}}} \right) \cr}
\)
dinsdag 9 juni 2020
Nagekomen bericht:-)
\(
\begin{array}{l}
y'' - y' - 2y = 0 \\
y( - 1) = 1 \\
y(1) = 0 \\
k^2 - k - 2 = 0 \\
(k - 2)(k + 1) = 0 \\
k = 2 \vee k = - 1 \\
y = C_1 \cdot e^{2x} + C_2 \cdot e^{ - x} \\
\left\{ \begin{array}{l}
C_1 \cdot e^{ - 2} + C_2 \cdot e = 1 \\
C_1 \cdot e^2 + C_2 \cdot e^{ - 1} = 0 \\
\end{array} \right. \\
\left\{ \begin{array}{l}
C_1 = \frac{{e^2 }}{{1 - e^6 }} \\
C_2 = \frac{{e^5 }}{{e^6 - 1}} \\
\end{array} \right. \\
y = \frac{{e^2 }}{{1 - e^6 }} \cdot e^{2x} + \frac{{e^5 }}{{e^6 - 1}} \cdot e^{ - x} \\
y = - \frac{{e^2 }}{{e^6 - 1}} \cdot e^{2x} + \frac{{e^5 }}{{e^6 - 1}} \cdot e^{ - x} \\
y = \frac{{e^{5 - x} - e^{2x + 2} }}{{e^6 - 1}} \\
\end{array}
\)
maandag 8 juni 2020
Quotientregel
\(
\eqalign{
& \left[ {\frac{t}
{n}} \right]' = \cr
& \left[ {t \cdot n^{ - 1} } \right]' = \cr
& t' \cdot n^{ - 1} + t \cdot - 1 \cdot n^{ - 2} \cdot n' = \cr
& \frac{{t'}}
{n} - \frac{{t \cdot n'}}
{{n^2 }} = \cr
& \frac{{t' \cdot n}}
{{n^2 }} - \frac{{t \cdot n'}}
{{n^2 }} = \cr
& \frac{{t' \cdot n - t \cdot n'}}
{{n^2 }} \cr}
\)
vrijdag 5 juni 2020
De som
8.1 Bereken de som van de volgende rijen getallen.
a. 2007006
b. 494550
c. 750000
d. 49800
e. 9899100
f. 9902700
De som van een rekenkundige rij gaat prima met:
Opmerking
\(
som = \frac{1}{2} \cdot 1800 \cdot \left( {1006 + 9997} \right)
\)
...en dan maar hopen dat het klopt...😏
- Alle positieve gehele getallen van 1 tot en met 2003.
- Alle positieve gehele getallen van drie cijfers.
- Alle oneven getallen tussen 1000 en 2000.
- Alle positieve gehele getallen van hoogstens drie cijfers die op het cijfer 3 eindigen.
- Alle positieve gehele getallen van vier cijfers die eindigen op het cijfer 2 of het cijfer 7.
- Alle positieve gehele getallen van vier cijfers die eindigen op het cijfer 6 of het cijfer 7.
a. 2007006
b. 494550
c. 750000
d. 49800
e. 9899100
f. 9902700
De som van een rekenkundige rij gaat prima met:
- som=1/2·aantal termen·(eerste term+laatste term)
- rekenkundige rijen
Opmerking
\(
som = \frac{1}{2} \cdot 1800 \cdot \left( {1006 + 9997} \right)
\)
...en dan maar hopen dat het klopt...😏
woensdag 3 juni 2020
Hoe moeilijk kan dat zijn?:-)
\(
\eqalign{
& \int {{{\tan (\ln (\sqrt x ))} \over x}} \,dx = \cr
& \int {\tan \left( {{1 \over 2}\ln (x)} \right)} \cdot {1 \over x}\,dx = \cr
& \int {\tan \left( {{1 \over 2}\ln (x)} \right)} \,d\left( {\ln (x)} \right) = \cr
& \downarrow u = \ln (x) \cr
& \int {\tan \left( {{1 \over 2}u} \right)} \,du = \cr
& \int {{{\sin \left( {{1 \over 2}u} \right)} \over {\cos \left( {{1 \over 2}u} \right)}}} \,du = \cr
& \int {{{ - 2} \over {\cos \left( {{1 \over 2}u} \right)}}} \cdot - {1 \over 2}\sin \left( {{1 \over 2}u} \right)\,du = \cr
& \int {{{ - 2} \over {\cos \left( {{1 \over 2}u} \right)}}} \cdot \,d\left( {\cos \left( {{1 \over 2}u} \right)} \right) = \cr
& \downarrow v = \cos \left( {{1 \over 2}u} \right) \cr
& \int {{{ - 2} \over v}} \cdot \,dv = \cr
& - 2\ln (v) + C = \cr
& - 2\ln \left( {\cos \left( {{1 \over 2}u} \right)} \right) + C = \cr
& - 2\ln \left( {\cos \left( {{1 \over 2}\ln (x)} \right)} \right) + C \cr
& - 2\ln \left( {\cos \left( {\ln (\sqrt x )} \right)} \right) + C \cr}
\)
\eqalign{
& \int {{{\tan (\ln (\sqrt x ))} \over x}} \,dx = \cr
& \int {\tan \left( {{1 \over 2}\ln (x)} \right)} \cdot {1 \over x}\,dx = \cr
& \int {\tan \left( {{1 \over 2}\ln (x)} \right)} \,d\left( {\ln (x)} \right) = \cr
& \downarrow u = \ln (x) \cr
& \int {\tan \left( {{1 \over 2}u} \right)} \,du = \cr
& \int {{{\sin \left( {{1 \over 2}u} \right)} \over {\cos \left( {{1 \over 2}u} \right)}}} \,du = \cr
& \int {{{ - 2} \over {\cos \left( {{1 \over 2}u} \right)}}} \cdot - {1 \over 2}\sin \left( {{1 \over 2}u} \right)\,du = \cr
& \int {{{ - 2} \over {\cos \left( {{1 \over 2}u} \right)}}} \cdot \,d\left( {\cos \left( {{1 \over 2}u} \right)} \right) = \cr
& \downarrow v = \cos \left( {{1 \over 2}u} \right) \cr
& \int {{{ - 2} \over v}} \cdot \,dv = \cr
& - 2\ln (v) + C = \cr
& - 2\ln \left( {\cos \left( {{1 \over 2}u} \right)} \right) + C = \cr
& - 2\ln \left( {\cos \left( {{1 \over 2}\ln (x)} \right)} \right) + C \cr
& - 2\ln \left( {\cos \left( {\ln (\sqrt x )} \right)} \right) + C \cr}
\)
zondag 17 mei 2020
Hoe moeilijk kan dat zijn?:-)
Hoe bereken je de eerste term van een meetkundige rij als je weet dat t6+t7=240 en t9+t10=1300?
Uitwerking
\( \eqalign{ & \left\{ \matrix{ ar^5 + ar^6 = 240 \hfill \cr ar^8 + ar^9 = 1300 \hfill \cr} \right. \cr & {{ar^8 + ar^9 } \over {ar^5 + ar^6 }} = {{1300} \over {240}} \cr & {{ar^8 \left( {1 + r} \right)} \over {ar^5 \left( {1 + r} \right)}} = {{65} \over {12}} \cr & r^3 = {{65} \over {12}} \cr & r = {1 \over 6}\root 3 \of {1170} \cr & a \cdot \left( {{1 \over 6}\root 3 \of {1170} } \right)^5 + a \cdot \left( {{1 \over 6}\root 3 \of {1170} } \right)^6 = 240 \cr & a\left( {\left( {{1 \over 6}\root 3 \of {1170} } \right)^5 + \left( {{1 \over 6}\root 3 \of {1170} } \right)^6 } \right) = 240 \cr & a = {{240} \over {\left( {{1 \over 6}\root 3 \of {1170} } \right)^5 + \left( {{1 \over 6}\root 3 \of {1170} } \right)^6 }} \cr & a = {{240} \over {{{65\root 3 \of {50700} } \over {144}} + {{4225} \over {144}}}} \cr & a = {{34560} \over {65\root 3 \of {50700} + 4225}} \cr & a = {{6912} \over {13\root 3 \of {50700} + 845}} \cr & a = {{6912} \over {13\root 3 \of {50700} + 845}} \cdot {{\root 3 \of {65} } \over {\root 3 \of {65} }} \cr & a = {{6912\root 3 \of {65} } \over {845\root 3 \of {12} + 845\root 3 \of {65} }} \cr & a = {{6912\root 3 \of {65} } \over {845\left( {\root 3 \of {65} + \root 3 \of {12} } \right)}} \cr} \)
Uitwerking
\( \eqalign{ & \left\{ \matrix{ ar^5 + ar^6 = 240 \hfill \cr ar^8 + ar^9 = 1300 \hfill \cr} \right. \cr & {{ar^8 + ar^9 } \over {ar^5 + ar^6 }} = {{1300} \over {240}} \cr & {{ar^8 \left( {1 + r} \right)} \over {ar^5 \left( {1 + r} \right)}} = {{65} \over {12}} \cr & r^3 = {{65} \over {12}} \cr & r = {1 \over 6}\root 3 \of {1170} \cr & a \cdot \left( {{1 \over 6}\root 3 \of {1170} } \right)^5 + a \cdot \left( {{1 \over 6}\root 3 \of {1170} } \right)^6 = 240 \cr & a\left( {\left( {{1 \over 6}\root 3 \of {1170} } \right)^5 + \left( {{1 \over 6}\root 3 \of {1170} } \right)^6 } \right) = 240 \cr & a = {{240} \over {\left( {{1 \over 6}\root 3 \of {1170} } \right)^5 + \left( {{1 \over 6}\root 3 \of {1170} } \right)^6 }} \cr & a = {{240} \over {{{65\root 3 \of {50700} } \over {144}} + {{4225} \over {144}}}} \cr & a = {{34560} \over {65\root 3 \of {50700} + 4225}} \cr & a = {{6912} \over {13\root 3 \of {50700} + 845}} \cr & a = {{6912} \over {13\root 3 \of {50700} + 845}} \cdot {{\root 3 \of {65} } \over {\root 3 \of {65} }} \cr & a = {{6912\root 3 \of {65} } \over {845\root 3 \of {12} + 845\root 3 \of {65} }} \cr & a = {{6912\root 3 \of {65} } \over {845\left( {\root 3 \of {65} + \root 3 \of {12} } \right)}} \cr} \)
donderdag 14 mei 2020
Een assenvergelijking van het vlak V
- Wat is de vergelijking voor vlak V met de punten E, P en Q?
\( \eqalign{ & V:(8,0,0),(0,4,0)\,\,\,en\,\,\,(0,0, - 8) \cr & V:\frac{x} {8} + \frac{y} {4} + \frac{z} {{ - 8}} = 1 \cr & V:x + 2y - z = 8 \cr} \)
zaterdag 25 april 2020
Snijlijn van twee vlakken
In een rechthoekig assenstelsel OXYZ zijn gegeven de punten A(4,0,0), B(4,8,0), C (0,8,0) en T (0,0,8). Op AB ligt punt D(4,6,0).
\( \begin{array}{l} TDO:\lambda \left( {\begin{array}{*{20}c} 2 \\ 3 \\ 0 \\ \end{array}} \right) + \mu \left( {\begin{array}{*{20}c} 0 \\ 0 \\ 1 \\ \end{array}} \right) \\ ACT:\left( {\begin{array}{*{20}c} 4 \\ 0 \\ 0 \\ \end{array}} \right) + \rho \left( {\begin{array}{*{20}c} { - 1} \\ 2 \\ 0 \\ \end{array}} \right) + \tau \left( {\begin{array}{*{20}c} { - 1} \\ 0 \\ 2 \\ \end{array}} \right) \\ \left\{ \begin{array}{l} 2\lambda = 4 - \rho - \tau \\ 3\lambda = 2\rho \\ \mu = 2\tau \\ \end{array} \right. \\ \left\{ \begin{array}{l} 4\lambda = 8 - 2\rho - 2\tau \\ 3\lambda = 2\rho \\ \mu = 2\tau \\ \end{array} \right. \\ \left\{ \begin{array}{l} 4\lambda = 8 - 3\lambda - \mu \\ 3\lambda = 2\rho \\ \mu = 2\tau \\ \end{array} \right. \\ \mu = - 7\lambda + 8 \\ l:\lambda \left( {\begin{array}{*{20}c} 2 \\ 3 \\ 0 \\ \end{array}} \right) + \left( { - 7\lambda + 8} \right)\left( {\begin{array}{*{20}c} 0 \\ 0 \\ 1 \\ \end{array}} \right) \\ l:\lambda \left( {\begin{array}{*{20}c} 2 \\ 3 \\ 0 \\ \end{array}} \right) + - 7\lambda \cdot \left( {\begin{array}{*{20}c} 0 \\ 0 \\ 1 \\ \end{array}} \right) + 8 \cdot \left( {\begin{array}{*{20}c} 0 \\ 0 \\ 1 \\ \end{array}} \right) \\ l:\left( {\begin{array}{*{20}c} 0 \\ 0 \\ 8 \\ \end{array}} \right) + \lambda \left( {\begin{array}{*{20}c} 2 \\ 3 \\ { - 7} \\ \end{array}} \right) \\ \end{array} \)
- Bepaal een vectorvoorstelling van de snijlijn van vlak TDO en vlak ACT.
\( \begin{array}{l} TDO:\lambda \left( {\begin{array}{*{20}c} 2 \\ 3 \\ 0 \\ \end{array}} \right) + \mu \left( {\begin{array}{*{20}c} 0 \\ 0 \\ 1 \\ \end{array}} \right) \\ ACT:\left( {\begin{array}{*{20}c} 4 \\ 0 \\ 0 \\ \end{array}} \right) + \rho \left( {\begin{array}{*{20}c} { - 1} \\ 2 \\ 0 \\ \end{array}} \right) + \tau \left( {\begin{array}{*{20}c} { - 1} \\ 0 \\ 2 \\ \end{array}} \right) \\ \left\{ \begin{array}{l} 2\lambda = 4 - \rho - \tau \\ 3\lambda = 2\rho \\ \mu = 2\tau \\ \end{array} \right. \\ \left\{ \begin{array}{l} 4\lambda = 8 - 2\rho - 2\tau \\ 3\lambda = 2\rho \\ \mu = 2\tau \\ \end{array} \right. \\ \left\{ \begin{array}{l} 4\lambda = 8 - 3\lambda - \mu \\ 3\lambda = 2\rho \\ \mu = 2\tau \\ \end{array} \right. \\ \mu = - 7\lambda + 8 \\ l:\lambda \left( {\begin{array}{*{20}c} 2 \\ 3 \\ 0 \\ \end{array}} \right) + \left( { - 7\lambda + 8} \right)\left( {\begin{array}{*{20}c} 0 \\ 0 \\ 1 \\ \end{array}} \right) \\ l:\lambda \left( {\begin{array}{*{20}c} 2 \\ 3 \\ 0 \\ \end{array}} \right) + - 7\lambda \cdot \left( {\begin{array}{*{20}c} 0 \\ 0 \\ 1 \\ \end{array}} \right) + 8 \cdot \left( {\begin{array}{*{20}c} 0 \\ 0 \\ 1 \\ \end{array}} \right) \\ l:\left( {\begin{array}{*{20}c} 0 \\ 0 \\ 8 \\ \end{array}} \right) + \lambda \left( {\begin{array}{*{20}c} 2 \\ 3 \\ { - 7} \\ \end{array}} \right) \\ \end{array} \)
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