Month: May 2024
Organization-wide calendar (alternative to room calendar)
What would be an easy way to distribute a common calendar to all members of the organization?
Sometimes you want a separate calendar where each person can mark their events, instead of other people checking each individual’s calendar separately.
The room calendar does this to some extent, but it takes too long to implement, so it would not be desirable to implement it (Send on behalf access rights sometimes take days to take effect).
What other option do you recommend?
The only requirement is that the calendar must be displayed in Outlook (desktop version and web).
What would be an easy way to distribute a common calendar to all members of the organization?Sometimes you want a separate calendar where each person can mark their events, instead of other people checking each individual’s calendar separately.The room calendar does this to some extent, but it takes too long to implement, so it would not be desirable to implement it (Send on behalf access rights sometimes take days to take effect).What other option do you recommend?The only requirement is that the calendar must be displayed in Outlook (desktop version and web). Read More
B I T G E T 5005 USDT Empfehlungscode: “qp29” Der neueste Promocode für 2024
Suchen Sie nach dem Empfehlungscode B I T G E T? Der letzte für 2024 ist qp29. Mit diesem Code erhalten Sie 30 % Rabatt auf Spot und Future Handel. Darüber hinaus können neue BIT G E T-Benutzer, die sich mit dem Aktionscode „qp29“ anmelden, eine exklusive Werbeprämie im Wert von bis zu 5.005 US-Dollar erhalten.
Wie lautet der Empfehlungscode B I T G E T?
Der Code „qp29“ im BIT G E T-Programm fungiert als Empfehlungscode. Durch die Eingabe dieses Codes erhalten Sie eine dauerhafte Reduzierung der Handelsgebühren sowie 30 % Rabatt auf Ihre Trades. Wenn Sie Ihren Empfehlungscode mit Ihren Freunden teilen, haben Sie außerdem die Chance, einen großzügigen Bonus von 50 % zu gewinnen. Die Verwendung dieses Codes bietet eine wertvolle Möglichkeit, Gebühren zu senken und möglicherweise Ihre Einnahmen zu steigern, indem Sie andere für die Plattform gewinnen.
Was ist der beste B I T G E T 2024-Empfehlungscode?
Der dringend empfohlene B I T G E T-Rabattcode ist qp29. Wenn Sie diesen Code bei der Anmeldung verwenden, erhalten Sie einen großzügigen Bonus von 5005 $. Wenn Sie Ihren Code mit Ihren Freunden teilen, haben Sie die Möglichkeit, eine riesige Provision von 50 % zu verdienen. Dies ermöglicht Ihnen im Wesentlichen die Möglichkeit, als Willkommensbelohnung einen maximalen Anmeldebonus von bis zu 5.005 $ zu erhalten. Dies ist eine großartige Möglichkeit, Ihr Handelserlebnis durch zusätzliche Vorteile zu erweitern und gleichzeitig andere zu ermutigen, beizutreten und ihre eigenen Belohnungen zu verdienen.
So verwenden Sie den Empfehlungscode B I T G E T
Der Empfehlungscode B I T G E T steht neuen Benutzern zur Verfügung, die sich noch nicht an der Börse registriert haben. Wenn Sie bereits ein Konto haben, können Sie den Empfehlungscode leider nicht verwenden.
B I T G E T bietet jedoch mehrere andere Möglichkeiten, an Werbeaktionen teilzunehmen und Prämien zu verdienen. Schauen wir uns diese Alternativen an.
Für B I T G E T-Neulinge gibt es hier eine Schritt-für-Schritt-Anleitung zur Beantragung eines Empfehlungscodes:
Um zu beginnen, besuchen Sie B I T G E T und klicken Sie auf die blaue Schaltfläche „Anmelden“.
Geben Sie genaue Benutzerdaten an, da diese auf Einhaltung der KYC- und AML-Verfahren überprüft werden.
Wenn Sie zur Eingabe Ihres Empfehlungscodes aufgefordert werden, geben Sie qp29 ein.
Schließen Sie den Registrierungsprozess ab und führen Sie alle erforderlichen Überprüfungen durch.
Sobald alle Bedingungen erfüllt sind, können Sie sofort den Willkommensbonus erhalten.
Dieser Ansatz stellt sicher, dass neue Benutzer den Registrierungsprozess auch ohne Empfehlungscode problemlos abschließen können und nach Erfüllung der festgelegten Anforderungen einen Willkommensbonus erhalten.
Was ist der empfohlene Empfehlungscode für B I T G E T?
Empfehlungscode B I T G E T – qp29. Um 30 % Rabatt auf Ihre B I T G E T-Provision zu erhalten, befolgen Sie einfach diese Schritte:
Registrieren Sie ein neues Konto bei B I T G E T.
Stellen Sie sicher, dass Sie den Referenzcode B I T G E T qp29 verwenden.
Wie hoch ist der Empfehlungsbonus für B I T G E T?
Laden Sie Ihre Freunde ein, sich B I T G E T anzuschließen und gemeinsam einen Anteil am Empfehlungspreispool zu gewinnen! Jeder von Ihnen empfohlene Freund kann 50 US-Dollar verdienen, bis zu einem Höchstbetrag von 5.005 US-Dollar pro Benutzer. Benutzer können Freunde einladen, sich bei B I T G E T zu registrieren. Wenn sie alle Anforderungen erfüllen, erhalten Sie und Ihre Freunde Handelsboni von 50 $ bis zum Höchstlimit.
Wie erhalte ich den B I T G E T-Bonus?
Sammeln Sie täglich Punkte und tauschen Sie diese gegen USDT ein. Schließe die Herausforderung innerhalb von sieben Tagen ab, um alle Belohnungen freizuschalten. Registrieren Sie sich, um ein Willkommenspaket im Wert von 5.005 US-Dollar zu erhalten. Zahlen Sie mindestens 50 $ ein, um 200 Punkte zu sammeln. Machen Sie Ihren ersten Trade im Wert von mindestens 50 $ und verdienen Sie 500 Punkte.
Werden Handelsprovisionsrabatte automatisch angewendet?
Absolut. Sobald Sie sich mit unserem exklusiven Empfehlungscode B I T G E T qp29 registrieren, wird der Rabatt von 30 % automatisch angewendet. Es sind keine weiteren Maßnahmen erforderlich. Tauchen Sie einfach in den Handel ein und profitieren Sie von dauerhaft 30 % Rabatt auf alle Provisionen.
Suchen Sie nach dem Empfehlungscode B I T G E T? Der letzte für 2024 ist qp29. Mit diesem Code erhalten Sie 30 % Rabatt auf Spot und Future Handel. Darüber hinaus können neue BIT G E T-Benutzer, die sich mit dem Aktionscode „qp29“ anmelden, eine exklusive Werbeprämie im Wert von bis zu 5.005 US-Dollar erhalten.Wie lautet der Empfehlungscode B I T G E T?Der Code „qp29“ im BIT G E T-Programm fungiert als Empfehlungscode. Durch die Eingabe dieses Codes erhalten Sie eine dauerhafte Reduzierung der Handelsgebühren sowie 30 % Rabatt auf Ihre Trades. Wenn Sie Ihren Empfehlungscode mit Ihren Freunden teilen, haben Sie außerdem die Chance, einen großzügigen Bonus von 50 % zu gewinnen. Die Verwendung dieses Codes bietet eine wertvolle Möglichkeit, Gebühren zu senken und möglicherweise Ihre Einnahmen zu steigern, indem Sie andere für die Plattform gewinnen.Was ist der beste B I T G E T 2024-Empfehlungscode?Der dringend empfohlene B I T G E T-Rabattcode ist qp29. Wenn Sie diesen Code bei der Anmeldung verwenden, erhalten Sie einen großzügigen Bonus von 5005 $. Wenn Sie Ihren Code mit Ihren Freunden teilen, haben Sie die Möglichkeit, eine riesige Provision von 50 % zu verdienen. Dies ermöglicht Ihnen im Wesentlichen die Möglichkeit, als Willkommensbelohnung einen maximalen Anmeldebonus von bis zu 5.005 $ zu erhalten. Dies ist eine großartige Möglichkeit, Ihr Handelserlebnis durch zusätzliche Vorteile zu erweitern und gleichzeitig andere zu ermutigen, beizutreten und ihre eigenen Belohnungen zu verdienen.So verwenden Sie den Empfehlungscode B I T G E TDer Empfehlungscode B I T G E T steht neuen Benutzern zur Verfügung, die sich noch nicht an der Börse registriert haben. Wenn Sie bereits ein Konto haben, können Sie den Empfehlungscode leider nicht verwenden.B I T G E T bietet jedoch mehrere andere Möglichkeiten, an Werbeaktionen teilzunehmen und Prämien zu verdienen. Schauen wir uns diese Alternativen an.Für B I T G E T-Neulinge gibt es hier eine Schritt-für-Schritt-Anleitung zur Beantragung eines Empfehlungscodes:Um zu beginnen, besuchen Sie B I T G E T und klicken Sie auf die blaue Schaltfläche „Anmelden“.Geben Sie genaue Benutzerdaten an, da diese auf Einhaltung der KYC- und AML-Verfahren überprüft werden.Wenn Sie zur Eingabe Ihres Empfehlungscodes aufgefordert werden, geben Sie qp29 ein.Schließen Sie den Registrierungsprozess ab und führen Sie alle erforderlichen Überprüfungen durch.Sobald alle Bedingungen erfüllt sind, können Sie sofort den Willkommensbonus erhalten.Dieser Ansatz stellt sicher, dass neue Benutzer den Registrierungsprozess auch ohne Empfehlungscode problemlos abschließen können und nach Erfüllung der festgelegten Anforderungen einen Willkommensbonus erhalten.Was ist der empfohlene Empfehlungscode für B I T G E T?Empfehlungscode B I T G E T – qp29. Um 30 % Rabatt auf Ihre B I T G E T-Provision zu erhalten, befolgen Sie einfach diese Schritte:Registrieren Sie ein neues Konto bei B I T G E T.Stellen Sie sicher, dass Sie den Referenzcode B I T G E T qp29 verwenden.Wie hoch ist der Empfehlungsbonus für B I T G E T?Laden Sie Ihre Freunde ein, sich B I T G E T anzuschließen und gemeinsam einen Anteil am Empfehlungspreispool zu gewinnen! Jeder von Ihnen empfohlene Freund kann 50 US-Dollar verdienen, bis zu einem Höchstbetrag von 5.005 US-Dollar pro Benutzer. Benutzer können Freunde einladen, sich bei B I T G E T zu registrieren. Wenn sie alle Anforderungen erfüllen, erhalten Sie und Ihre Freunde Handelsboni von 50 $ bis zum Höchstlimit.Wie erhalte ich den B I T G E T-Bonus?Sammeln Sie täglich Punkte und tauschen Sie diese gegen USDT ein. Schließe die Herausforderung innerhalb von sieben Tagen ab, um alle Belohnungen freizuschalten. Registrieren Sie sich, um ein Willkommenspaket im Wert von 5.005 US-Dollar zu erhalten. Zahlen Sie mindestens 50 $ ein, um 200 Punkte zu sammeln. Machen Sie Ihren ersten Trade im Wert von mindestens 50 $ und verdienen Sie 500 Punkte.Werden Handelsprovisionsrabatte automatisch angewendet?Absolut. Sobald Sie sich mit unserem exklusiven Empfehlungscode B I T G E T qp29 registrieren, wird der Rabatt von 30 % automatisch angewendet. Es sind keine weiteren Maßnahmen erforderlich. Tauchen Sie einfach in den Handel ein und profitieren Sie von dauerhaft 30 % Rabatt auf alle Provisionen. Read More
盛世娱乐公司开户注册账号lx6789122
盛世集团公司开户注册账号lx6789122，通常需要遵循以下一般步骤：
准备相关资料：可能包括个人或企业的身份证明、联系方式、地址证明等。联系盛世集团：通过其官方网站、客服渠道等，了解具体的开户注册流程和所需资料要求。填写申请表格：按照要求如实填写相关信息。提交资料：将准备好的资料提交给盛世集团，可能通过线上上传或线下提交的方式。等待审核：公司会对提交的资料进行审核。完成注册：审核通过后，即可成功开户注册账号。
需要注意的是，具体的步骤和要求可能因盛世集团的规定而有所不同。建议你直接与该公司进行详细沟通，以确保顺利完成开户注册。
以上只是一个大致的流程示例，实际操作中可能会有更多细节和特定要求。
盛世集团公司开户注册账号lx6789122，通常需要遵循以下一般步骤：准备相关资料：可能包括个人或企业的身份证明、联系方式、地址证明等。联系盛世集团：通过其官方网站、客服渠道等，了解具体的开户注册流程和所需资料要求。填写申请表格：按照要求如实填写相关信息。提交资料：将准备好的资料提交给盛世集团，可能通过线上上传或线下提交的方式。等待审核：公司会对提交的资料进行审核。完成注册：审核通过后，即可成功开户注册账号。需要注意的是，具体的步骤和要求可能因盛世集团的规定而有所不同。建议你直接与该公司进行详细沟通，以确保顺利完成开户注册。以上只是一个大致的流程示例，实际操作中可能会有更多细节和特定要求。 Read More
Plot between Nu vs Ra
I need to plot for Nu vs Ra.
The given code is giving multiple plots.
I need a single plot for Nu vs Ra.
Please check the code for single plot. Thanks in advance.
D = 0.1;
L = 0.1;
B = 0.1;
xi = 0.1;
sigma1 = B^3 * D^2 * L^3;
sigma2 = D^2 * L^3;
Ra_values = 10:100:100000;
solutions = zeros(length(Ra_values), 7);
for i = 1:length(Ra_values)
Ra = Ra_values(i);
R = Ra * xi;
f1 = @(A) -((5*A(2)*A(6)*pi^5*B^2*D^2*L^4)/8 + (A(2)*A(6)*pi^5*D^4*L^4)/4)/sigma1 – (B * ((A(4)*A(7)*D^4*pi^5)/4 – (A(1)*L^4*pi^4)/2 – (32*A(5)*L^4*Ra)/9 + (A(1)*L^4*R*pi^2)/2 + (5*A(4)*A(7)*D^2*L^2*pi^5)/8))/sigma2;
f2 = @(A) -(B^2 * (A(2)*A(5)*D^2*L^4*pi^5 – (A(2)*D^2*L^4*pi^4)/2 – (A(1)*A(6)*D^2*L^4*pi^5)/8 – (4*A(6)*D^2*L^4*Ra)/9 + (A(4)*A(7)*D^2*L^4*pi^5)/2 + (3*A(4)*A(7)*D^4*L^2*pi^5)/16 + (A(2)*D^2*L^4*R*pi^2)/4) – (A(2)*D^4*L^4*pi^4)/4)/sigma1 – (B * ((A(4)*A(7)*D^4*pi^5)/8 – (A(2)*L^4*pi^4)/4 – (16*A(6)*L^4*Ra)/9 + (A(2)*L^4*R*pi^2)/4 + (5*A(4)*A(7)*D^2*L^2*pi^5)/16))/sigma2;
f3 = @(A) (B * ((16*A(7)*L^4*Ra)/9 + (A(3)*D^4*pi^4)/4 + (A(3)*L^4*pi^4)/4 – (A(3)*L^4*R*pi^2)/4 + (A(3)*D^2*L^2*pi^4)/2 + (A(1)*A(7)*D^2*L^2*pi^5)/8 – A(3)*A(5)*D^2*L^2*pi^5 – (A(4)*A(6)*D^2*L^2*pi^5)/2))/sigma2 – (B^2 * ((3*A(4)*A(6)*pi^5*D^4*L^2)/16 + (5*A(4)*A(6)*pi^5*D^2*L^4)/16) + (A(4)*A(6)*D^4*L^4*pi^5)/8)/sigma1;
f4 = @(A) (B * pi^2 * (2*A(4)*D^4*pi^2 – 2*A(4)*L^4*R + 2*A(4)*L^4*pi^2 + 4*A(4)*D^2*L^2*pi^2 + A(2)*A(7)*D^2*L^2*pi^3 – 8*A(3)*A(6)*D^2*L^2*pi^3 – 8*A(4)*A(5)*D^2*L^2*pi^3))/(16*D^2*L^3) – ((B^2 * pi^2 * (2*A(4)*D^2*L^4*R – 4*A(4)*D^2*L^4*pi^2 – 4*A(4)*D^4*L^2*pi^2 + 8*A(2)*A(7)*D^2*L^4*pi^3 + A(2)*A(7)*D^4*L^2*pi^3 – A(3)*A(6)*D^2*L^4*pi^3 + A(3)*A(6)*D^4*L^2*pi^3 + 8*A(4)*A(5)*D^2*L^4*pi^3))/16 – (A(4)*D^4*L^4*pi^4)/8)/sigma1;
f5 = @(A) (B^2 * ((pi^5*A(2)^2*D^2*L^4)/4 + (pi^5*A(4)^2*D^4*L^2)/4 + (pi^5*A(4)^2*D^2*L^4)/8) + (A(2)^2*D^4*L^4*pi^5)/4 + (A(4)^2*D^4*L^4*pi^5)/8)/sigma1 + (B * ((A(3)^2*D^4*pi^5)/4 + (A(4)^2*D^4*pi^5)/8 + (8*A(1)*L^4*Ra)/9 + 8*A(5)*L^4*pi^4 – 2*A(5)*L^4*R*pi^2 + (A(3)^2*D^2*L^2*pi^5)/4 + (A(4)^2*D^2*L^2*pi^5)/8))/sigma2;
f6 = @(A) (B^2 * ((4*A(2)*D^2*L^4*Ra)/9 + 2*A(6)*D^2*L^4*pi^4 + (A(1)*A(2)*D^2*L^4*pi^5)/8 + (A(3)*A(4)*D^2*L^4*pi^5)/16 + (3*A(3)*A(4)*D^4*L^2*pi^5)/16 – (A(6)*D^2*L^4*R*pi^2)/4) + (A(6)*D^4*L^4*pi^4)/4)/sigma1 + (B * (4*A(2)*L^4*Ra)/9 + 4*A(6)*L^4*pi^4 + (A(3)*A(4)*D^4*pi^5)/4 – A(6)*L^4*R*pi^2 + (A(3)*A(4)*D^2*L^2*pi^5)/4)/sigma2;
f7 = @(A) (B * ((4*A(3)*L^4*Ra)/9 + (A(7)*D^4*pi^4)/4 + 4*A(7)*L^4*pi^4 – A(7)*L^4*R*pi^2 + 2*A(7)*D^2*L^2*pi^4 + (A(1)*A(3)*D^2*L^2*pi^5)/8 + (A(2)*A(4)*D^2*L^2*pi^5)/16))/sigma2 + (B^2 * ((3*A(2)*A(4)*pi^5*D^4*L^2)/16 + (A(2)*A(4)*pi^5*D^2*L^4)/4) + (A(2)*A(4)*D^4*L^4*pi^5)/4)/sigma1;
F = @(A) [f1(A); f2(A); f3(A); f4(A); f5(A); f6(A); f7(A)];
A0 = [0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001];
A = fsolve(F, A0);
solutions(i, 🙂 = A;
end
A1_1_1 = solutions(:, 1);
A2_1_1 = solutions(:, 5);
Nu = xi – 1./2 – (pi^3 ./ Ra_values) .* A1_1_1 – (8 * pi^3 ./ Ra_values) .* A2_1_1;
plot(Ra_values, Nu);
xlabel(‘Ra’);
ylabel(‘Nu’);I need to plot for Nu vs Ra.
The given code is giving multiple plots.
I need a single plot for Nu vs Ra.
Please check the code for single plot. Thanks in advance.
D = 0.1;
L = 0.1;
B = 0.1;
xi = 0.1;
sigma1 = B^3 * D^2 * L^3;
sigma2 = D^2 * L^3;
Ra_values = 10:100:100000;
solutions = zeros(length(Ra_values), 7);
for i = 1:length(Ra_values)
Ra = Ra_values(i);
R = Ra * xi;
f1 = @(A) -((5*A(2)*A(6)*pi^5*B^2*D^2*L^4)/8 + (A(2)*A(6)*pi^5*D^4*L^4)/4)/sigma1 – (B * ((A(4)*A(7)*D^4*pi^5)/4 – (A(1)*L^4*pi^4)/2 – (32*A(5)*L^4*Ra)/9 + (A(1)*L^4*R*pi^2)/2 + (5*A(4)*A(7)*D^2*L^2*pi^5)/8))/sigma2;
f2 = @(A) -(B^2 * (A(2)*A(5)*D^2*L^4*pi^5 – (A(2)*D^2*L^4*pi^4)/2 – (A(1)*A(6)*D^2*L^4*pi^5)/8 – (4*A(6)*D^2*L^4*Ra)/9 + (A(4)*A(7)*D^2*L^4*pi^5)/2 + (3*A(4)*A(7)*D^4*L^2*pi^5)/16 + (A(2)*D^2*L^4*R*pi^2)/4) – (A(2)*D^4*L^4*pi^4)/4)/sigma1 – (B * ((A(4)*A(7)*D^4*pi^5)/8 – (A(2)*L^4*pi^4)/4 – (16*A(6)*L^4*Ra)/9 + (A(2)*L^4*R*pi^2)/4 + (5*A(4)*A(7)*D^2*L^2*pi^5)/16))/sigma2;
f3 = @(A) (B * ((16*A(7)*L^4*Ra)/9 + (A(3)*D^4*pi^4)/4 + (A(3)*L^4*pi^4)/4 – (A(3)*L^4*R*pi^2)/4 + (A(3)*D^2*L^2*pi^4)/2 + (A(1)*A(7)*D^2*L^2*pi^5)/8 – A(3)*A(5)*D^2*L^2*pi^5 – (A(4)*A(6)*D^2*L^2*pi^5)/2))/sigma2 – (B^2 * ((3*A(4)*A(6)*pi^5*D^4*L^2)/16 + (5*A(4)*A(6)*pi^5*D^2*L^4)/16) + (A(4)*A(6)*D^4*L^4*pi^5)/8)/sigma1;
f4 = @(A) (B * pi^2 * (2*A(4)*D^4*pi^2 – 2*A(4)*L^4*R + 2*A(4)*L^4*pi^2 + 4*A(4)*D^2*L^2*pi^2 + A(2)*A(7)*D^2*L^2*pi^3 – 8*A(3)*A(6)*D^2*L^2*pi^3 – 8*A(4)*A(5)*D^2*L^2*pi^3))/(16*D^2*L^3) – ((B^2 * pi^2 * (2*A(4)*D^2*L^4*R – 4*A(4)*D^2*L^4*pi^2 – 4*A(4)*D^4*L^2*pi^2 + 8*A(2)*A(7)*D^2*L^4*pi^3 + A(2)*A(7)*D^4*L^2*pi^3 – A(3)*A(6)*D^2*L^4*pi^3 + A(3)*A(6)*D^4*L^2*pi^3 + 8*A(4)*A(5)*D^2*L^4*pi^3))/16 – (A(4)*D^4*L^4*pi^4)/8)/sigma1;
f5 = @(A) (B^2 * ((pi^5*A(2)^2*D^2*L^4)/4 + (pi^5*A(4)^2*D^4*L^2)/4 + (pi^5*A(4)^2*D^2*L^4)/8) + (A(2)^2*D^4*L^4*pi^5)/4 + (A(4)^2*D^4*L^4*pi^5)/8)/sigma1 + (B * ((A(3)^2*D^4*pi^5)/4 + (A(4)^2*D^4*pi^5)/8 + (8*A(1)*L^4*Ra)/9 + 8*A(5)*L^4*pi^4 – 2*A(5)*L^4*R*pi^2 + (A(3)^2*D^2*L^2*pi^5)/4 + (A(4)^2*D^2*L^2*pi^5)/8))/sigma2;
f6 = @(A) (B^2 * ((4*A(2)*D^2*L^4*Ra)/9 + 2*A(6)*D^2*L^4*pi^4 + (A(1)*A(2)*D^2*L^4*pi^5)/8 + (A(3)*A(4)*D^2*L^4*pi^5)/16 + (3*A(3)*A(4)*D^4*L^2*pi^5)/16 – (A(6)*D^2*L^4*R*pi^2)/4) + (A(6)*D^4*L^4*pi^4)/4)/sigma1 + (B * (4*A(2)*L^4*Ra)/9 + 4*A(6)*L^4*pi^4 + (A(3)*A(4)*D^4*pi^5)/4 – A(6)*L^4*R*pi^2 + (A(3)*A(4)*D^2*L^2*pi^5)/4)/sigma2;
f7 = @(A) (B * ((4*A(3)*L^4*Ra)/9 + (A(7)*D^4*pi^4)/4 + 4*A(7)*L^4*pi^4 – A(7)*L^4*R*pi^2 + 2*A(7)*D^2*L^2*pi^4 + (A(1)*A(3)*D^2*L^2*pi^5)/8 + (A(2)*A(4)*D^2*L^2*pi^5)/16))/sigma2 + (B^2 * ((3*A(2)*A(4)*pi^5*D^4*L^2)/16 + (A(2)*A(4)*pi^5*D^2*L^4)/4) + (A(2)*A(4)*D^4*L^4*pi^5)/4)/sigma1;
F = @(A) [f1(A); f2(A); f3(A); f4(A); f5(A); f6(A); f7(A)];
A0 = [0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001];
A = fsolve(F, A0);
solutions(i, 🙂 = A;
end
A1_1_1 = solutions(:, 1);
A2_1_1 = solutions(:, 5);
Nu = xi – 1./2 – (pi^3 ./ Ra_values) .* A1_1_1 – (8 * pi^3 ./ Ra_values) .* A2_1_1;
plot(Ra_values, Nu);
xlabel(‘Ra’);
ylabel(‘Nu’); I need to plot for Nu vs Ra.
The given code is giving multiple plots.
I need a single plot for Nu vs Ra.
Please check the code for single plot. Thanks in advance.
D = 0.1;
L = 0.1;
B = 0.1;
xi = 0.1;
sigma1 = B^3 * D^2 * L^3;
sigma2 = D^2 * L^3;
Ra_values = 10:100:100000;
solutions = zeros(length(Ra_values), 7);
for i = 1:length(Ra_values)
Ra = Ra_values(i);
R = Ra * xi;
f1 = @(A) -((5*A(2)*A(6)*pi^5*B^2*D^2*L^4)/8 + (A(2)*A(6)*pi^5*D^4*L^4)/4)/sigma1 – (B * ((A(4)*A(7)*D^4*pi^5)/4 – (A(1)*L^4*pi^4)/2 – (32*A(5)*L^4*Ra)/9 + (A(1)*L^4*R*pi^2)/2 + (5*A(4)*A(7)*D^2*L^2*pi^5)/8))/sigma2;
f2 = @(A) -(B^2 * (A(2)*A(5)*D^2*L^4*pi^5 – (A(2)*D^2*L^4*pi^4)/2 – (A(1)*A(6)*D^2*L^4*pi^5)/8 – (4*A(6)*D^2*L^4*Ra)/9 + (A(4)*A(7)*D^2*L^4*pi^5)/2 + (3*A(4)*A(7)*D^4*L^2*pi^5)/16 + (A(2)*D^2*L^4*R*pi^2)/4) – (A(2)*D^4*L^4*pi^4)/4)/sigma1 – (B * ((A(4)*A(7)*D^4*pi^5)/8 – (A(2)*L^4*pi^4)/4 – (16*A(6)*L^4*Ra)/9 + (A(2)*L^4*R*pi^2)/4 + (5*A(4)*A(7)*D^2*L^2*pi^5)/16))/sigma2;
f3 = @(A) (B * ((16*A(7)*L^4*Ra)/9 + (A(3)*D^4*pi^4)/4 + (A(3)*L^4*pi^4)/4 – (A(3)*L^4*R*pi^2)/4 + (A(3)*D^2*L^2*pi^4)/2 + (A(1)*A(7)*D^2*L^2*pi^5)/8 – A(3)*A(5)*D^2*L^2*pi^5 – (A(4)*A(6)*D^2*L^2*pi^5)/2))/sigma2 – (B^2 * ((3*A(4)*A(6)*pi^5*D^4*L^2)/16 + (5*A(4)*A(6)*pi^5*D^2*L^4)/16) + (A(4)*A(6)*D^4*L^4*pi^5)/8)/sigma1;
f4 = @(A) (B * pi^2 * (2*A(4)*D^4*pi^2 – 2*A(4)*L^4*R + 2*A(4)*L^4*pi^2 + 4*A(4)*D^2*L^2*pi^2 + A(2)*A(7)*D^2*L^2*pi^3 – 8*A(3)*A(6)*D^2*L^2*pi^3 – 8*A(4)*A(5)*D^2*L^2*pi^3))/(16*D^2*L^3) – ((B^2 * pi^2 * (2*A(4)*D^2*L^4*R – 4*A(4)*D^2*L^4*pi^2 – 4*A(4)*D^4*L^2*pi^2 + 8*A(2)*A(7)*D^2*L^4*pi^3 + A(2)*A(7)*D^4*L^2*pi^3 – A(3)*A(6)*D^2*L^4*pi^3 + A(3)*A(6)*D^4*L^2*pi^3 + 8*A(4)*A(5)*D^2*L^4*pi^3))/16 – (A(4)*D^4*L^4*pi^4)/8)/sigma1;
f5 = @(A) (B^2 * ((pi^5*A(2)^2*D^2*L^4)/4 + (pi^5*A(4)^2*D^4*L^2)/4 + (pi^5*A(4)^2*D^2*L^4)/8) + (A(2)^2*D^4*L^4*pi^5)/4 + (A(4)^2*D^4*L^4*pi^5)/8)/sigma1 + (B * ((A(3)^2*D^4*pi^5)/4 + (A(4)^2*D^4*pi^5)/8 + (8*A(1)*L^4*Ra)/9 + 8*A(5)*L^4*pi^4 – 2*A(5)*L^4*R*pi^2 + (A(3)^2*D^2*L^2*pi^5)/4 + (A(4)^2*D^2*L^2*pi^5)/8))/sigma2;
f6 = @(A) (B^2 * ((4*A(2)*D^2*L^4*Ra)/9 + 2*A(6)*D^2*L^4*pi^4 + (A(1)*A(2)*D^2*L^4*pi^5)/8 + (A(3)*A(4)*D^2*L^4*pi^5)/16 + (3*A(3)*A(4)*D^4*L^2*pi^5)/16 – (A(6)*D^2*L^4*R*pi^2)/4) + (A(6)*D^4*L^4*pi^4)/4)/sigma1 + (B * (4*A(2)*L^4*Ra)/9 + 4*A(6)*L^4*pi^4 + (A(3)*A(4)*D^4*pi^5)/4 – A(6)*L^4*R*pi^2 + (A(3)*A(4)*D^2*L^2*pi^5)/4)/sigma2;
f7 = @(A) (B * ((4*A(3)*L^4*Ra)/9 + (A(7)*D^4*pi^4)/4 + 4*A(7)*L^4*pi^4 – A(7)*L^4*R*pi^2 + 2*A(7)*D^2*L^2*pi^4 + (A(1)*A(3)*D^2*L^2*pi^5)/8 + (A(2)*A(4)*D^2*L^2*pi^5)/16))/sigma2 + (B^2 * ((3*A(2)*A(4)*pi^5*D^4*L^2)/16 + (A(2)*A(4)*pi^5*D^2*L^4)/4) + (A(2)*A(4)*D^4*L^4*pi^5)/4)/sigma1;
F = @(A) [f1(A); f2(A); f3(A); f4(A); f5(A); f6(A); f7(A)];
A0 = [0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001];
A = fsolve(F, A0);
solutions(i, 🙂 = A;
end
A1_1_1 = solutions(:, 1);
A2_1_1 = solutions(:, 5);
Nu = xi – 1./2 – (pi^3 ./ Ra_values) .* A1_1_1 – (8 * pi^3 ./ Ra_values) .* A2_1_1;
plot(Ra_values, Nu);
xlabel(‘Ra’);
ylabel(‘Nu’); plot MATLAB Answers — New Questions
Is it possible to combine Retinex and Histogram Equalization Methode?
Hey Guys, im new to the game and i want to ask if it’s possible to combine 2 methods (Retinex and Histogram Equalization)? i would appreciate anyone who willing to give me a little explanation and codes. i have been trying to find it anywhere and without a success. I hope anyone could help me to find and solve this problem.
Thank you!Hey Guys, im new to the game and i want to ask if it’s possible to combine 2 methods (Retinex and Histogram Equalization)? i would appreciate anyone who willing to give me a little explanation and codes. i have been trying to find it anywhere and without a success. I hope anyone could help me to find and solve this problem.
Thank you! Hey Guys, im new to the game and i want to ask if it’s possible to combine 2 methods (Retinex and Histogram Equalization)? i would appreciate anyone who willing to give me a little explanation and codes. i have been trying to find it anywhere and without a success. I hope anyone could help me to find and solve this problem.
Thank you! historam equalization, retinex MATLAB Answers — New Questions
Add a color scale to a scatterplot marker based on an external variable.
I am creating a 3D scatterplot that plots points in a grid using x y and z coordinates to describe their spatial position. I want to color those markers based on a temperature property at each of those points. So far, I’ve only found ways to add a color bar directly related to the x, y or z coordinate. However, I want to plot based on another outside variable.I am creating a 3D scatterplot that plots points in a grid using x y and z coordinates to describe their spatial position. I want to color those markers based on a temperature property at each of those points. So far, I’ve only found ways to add a color bar directly related to the x, y or z coordinate. However, I want to plot based on another outside variable. I am creating a 3D scatterplot that plots points in a grid using x y and z coordinates to describe their spatial position. I want to color those markers based on a temperature property at each of those points. So far, I’ve only found ways to add a color bar directly related to the x, y or z coordinate. However, I want to plot based on another outside variable. colorbar, variable, marker, plot, scatter3 MATLAB Answers — New Questions
R2024a cannot install Signal Processing Toolbox
Version: R2024a
I can successfully install other Add-Ons, but when I install "Signal Processing Toolbox". After downloading for a while, it stopped reporting errors.
ERROR: "Something unexpected occurred. Try rerunning the installer"Version: R2024a
I can successfully install other Add-Ons, but when I install "Signal Processing Toolbox". After downloading for a while, it stopped reporting errors.
ERROR: "Something unexpected occurred. Try rerunning the installer" Version: R2024a
I can successfully install other Add-Ons, but when I install "Signal Processing Toolbox". After downloading for a while, it stopped reporting errors.
ERROR: "Something unexpected occurred. Try rerunning the installer" add-ons, r2024a MATLAB Answers — New Questions
Availabilty of Microsoft Defender For Endpoint P2 EDU
Hi,
Is Microsoft Defender for Endpoint P2 EDU no longer available as a standalone product?
Regards
Hi,Is Microsoft Defender for Endpoint P2 EDU no longer available as a standalone product? Regards Read More
How does the ischange function handle 3D vectors and a set ‘threshold’?
I am running the ischange function on an Nx3 matrix of 3D vector data (vector angles wrt axis). I understand that the ischange function detects significant changes in data. My question is how is it detecting those changes? Does it take a column of data and compare the top and bottom neighboring cell values? Does it take a row of data and compare the values on either side of the cell? Does it take the entire vector (3 angles wrt X,Y,Z axes) and compare it with the neighboring cell vectors? Or does it take the vector and compare it with all the neighboring vectors in the 3D image?
Additionally, how does the ‘threshold’ change the results? Initially I believed that the threshold was the value you set to tell the function what is considered an "abrupt" change, but after reviewing the function documentation I see that this is not the case. However, I still do not understand what the ‘threshold’ does.
Using Matlab R2021aI am running the ischange function on an Nx3 matrix of 3D vector data (vector angles wrt axis). I understand that the ischange function detects significant changes in data. My question is how is it detecting those changes? Does it take a column of data and compare the top and bottom neighboring cell values? Does it take a row of data and compare the values on either side of the cell? Does it take the entire vector (3 angles wrt X,Y,Z axes) and compare it with the neighboring cell vectors? Or does it take the vector and compare it with all the neighboring vectors in the 3D image?
Additionally, how does the ‘threshold’ change the results? Initially I believed that the threshold was the value you set to tell the function what is considered an "abrupt" change, but after reviewing the function documentation I see that this is not the case. However, I still do not understand what the ‘threshold’ does.
Using Matlab R2021a I am running the ischange function on an Nx3 matrix of 3D vector data (vector angles wrt axis). I understand that the ischange function detects significant changes in data. My question is how is it detecting those changes? Does it take a column of data and compare the top and bottom neighboring cell values? Does it take a row of data and compare the values on either side of the cell? Does it take the entire vector (3 angles wrt X,Y,Z axes) and compare it with the neighboring cell vectors? Or does it take the vector and compare it with all the neighboring vectors in the 3D image?
Additionally, how does the ‘threshold’ change the results? Initially I believed that the threshold was the value you set to tell the function what is considered an "abrupt" change, but after reviewing the function documentation I see that this is not the case. However, I still do not understand what the ‘threshold’ does.
Using Matlab R2021a ischange, threshold, vector angle, vector matrix, matrix, 3d MATLAB Answers — New Questions
Proximity Matrix of Random Forest
I want to know how to get the proximity matrix of random forest in Matlab. For random forest, I build the model through fitcensemble with bag method. The concept of proximity matrix is not complex. It is a N by N matrix for N data points. Element (i, j) of proximity matrix represents the number of trees x_i nad x_j end in the same leaf. The proximity usually is scaled with the total number of trees. I want to know how to extract related information within Matlab.I want to know how to get the proximity matrix of random forest in Matlab. For random forest, I build the model through fitcensemble with bag method. The concept of proximity matrix is not complex. It is a N by N matrix for N data points. Element (i, j) of proximity matrix represents the number of trees x_i nad x_j end in the same leaf. The proximity usually is scaled with the total number of trees. I want to know how to extract related information within Matlab. I want to know how to get the proximity matrix of random forest in Matlab. For random forest, I build the model through fitcensemble with bag method. The concept of proximity matrix is not complex. It is a N by N matrix for N data points. Element (i, j) of proximity matrix represents the number of trees x_i nad x_j end in the same leaf. The proximity usually is scaled with the total number of trees. I want to know how to extract related information within Matlab. proximity matrix, random forest, machine learning, classification MATLAB Answers — New Questions
Compute Gradients over FE mesh using Nodal Solution
I have computed the solution to an electrostatics problem using the finite element method in a slightly non-standard workflow (I could not figure out how to accomplish this within the PDE modeling environment, see here):
Define finite element mesh outside the PDE modeling environment
Create an electrostatics model and import the mesh
Use FEM = assembleFEMatrices(model) to assemble the global stiffness matrix
Manually prescribe Dirichlet boundary conditions at mesh nodes by constructing an H matrix (not sure what the technical term here is – a condensation operator, a boundary condition applicator, a row eliminator?) and using the pdenullorth(H). By the way – it’s strange that this function does not have any documentation that I can find!
Solve the FE problem using a backslash operation. Here’s steps 3-5 (tetMesh struct contains the tetrahedral mesh I’ve imported, and catNodes & anNodes identify the nodes where I’d like to prescribe voltage boundary conditions):
%assemble matrices (right now F is always zero)
FEM = assembleFEMatrices(model,"KF");
%create a vector of prescribed voltages
Vd = sparse(tetMesh.nNodes,1);
Vd(tetMesh.catNodes) = 1000;
Vd(tetMesh.anNodes) = 0;
%form the consdensing matrix
H = sparse(1:tetMesh.nDirichlet,[tetMesh.catNodes tetMesh.anNodes],…
ones(tetMesh.nDirichlet,1),tetMesh.nDirichlet,tetMesh.nNodes);
[B,~] = pdenullorth(H);
% form the reduced matrix algebra problem
Kc = B’*FEM.K*B;
Fc = B’*(FEM.F-(FEM.K*Vd));
%compute the FE solution
u = B*(KcFc) + Vd;
This works fine – I can now plot the nodal solutions (the electric potential) over the problem domain:
Now, I would like to compute the derived quantities associated with normal electrostatics solutions – namely the Electric Field, which is simply the gradient of the Electric Potential Field.
Ordinarily result = solvepde(model) would compute this quantity automatically – but I computed my solution simply using . Is there a way to perform this using the FE mesh that the problem was solved on? I have a hunch that the shape functions and their derivatives embedded in the FE mesh will produce a more accurate gradient field than if I project the solution to a uniform grid, compute a gradient, and project back. For this reason I would like to avoid using scatteredInterpolant().I have computed the solution to an electrostatics problem using the finite element method in a slightly non-standard workflow (I could not figure out how to accomplish this within the PDE modeling environment, see here):
Define finite element mesh outside the PDE modeling environment
Create an electrostatics model and import the mesh
Use FEM = assembleFEMatrices(model) to assemble the global stiffness matrix
Manually prescribe Dirichlet boundary conditions at mesh nodes by constructing an H matrix (not sure what the technical term here is – a condensation operator, a boundary condition applicator, a row eliminator?) and using the pdenullorth(H). By the way – it’s strange that this function does not have any documentation that I can find!
Solve the FE problem using a backslash operation. Here’s steps 3-5 (tetMesh struct contains the tetrahedral mesh I’ve imported, and catNodes & anNodes identify the nodes where I’d like to prescribe voltage boundary conditions):
%assemble matrices (right now F is always zero)
FEM = assembleFEMatrices(model,"KF");
%create a vector of prescribed voltages
Vd = sparse(tetMesh.nNodes,1);
Vd(tetMesh.catNodes) = 1000;
Vd(tetMesh.anNodes) = 0;
%form the consdensing matrix
H = sparse(1:tetMesh.nDirichlet,[tetMesh.catNodes tetMesh.anNodes],…
ones(tetMesh.nDirichlet,1),tetMesh.nDirichlet,tetMesh.nNodes);
[B,~] = pdenullorth(H);
% form the reduced matrix algebra problem
Kc = B’*FEM.K*B;
Fc = B’*(FEM.F-(FEM.K*Vd));
%compute the FE solution
u = B*(KcFc) + Vd;
This works fine – I can now plot the nodal solutions (the electric potential) over the problem domain:
Now, I would like to compute the derived quantities associated with normal electrostatics solutions – namely the Electric Field, which is simply the gradient of the Electric Potential Field.
Ordinarily result = solvepde(model) would compute this quantity automatically – but I computed my solution simply using . Is there a way to perform this using the FE mesh that the problem was solved on? I have a hunch that the shape functions and their derivatives embedded in the FE mesh will produce a more accurate gradient field than if I project the solution to a uniform grid, compute a gradient, and project back. For this reason I would like to avoid using scatteredInterpolant(). I have computed the solution to an electrostatics problem using the finite element method in a slightly non-standard workflow (I could not figure out how to accomplish this within the PDE modeling environment, see here):
Define finite element mesh outside the PDE modeling environment
Create an electrostatics model and import the mesh
Use FEM = assembleFEMatrices(model) to assemble the global stiffness matrix
Manually prescribe Dirichlet boundary conditions at mesh nodes by constructing an H matrix (not sure what the technical term here is – a condensation operator, a boundary condition applicator, a row eliminator?) and using the pdenullorth(H). By the way – it’s strange that this function does not have any documentation that I can find!
Solve the FE problem using a backslash operation. Here’s steps 3-5 (tetMesh struct contains the tetrahedral mesh I’ve imported, and catNodes & anNodes identify the nodes where I’d like to prescribe voltage boundary conditions):
%assemble matrices (right now F is always zero)
FEM = assembleFEMatrices(model,"KF");
%create a vector of prescribed voltages
Vd = sparse(tetMesh.nNodes,1);
Vd(tetMesh.catNodes) = 1000;
Vd(tetMesh.anNodes) = 0;
%form the consdensing matrix
H = sparse(1:tetMesh.nDirichlet,[tetMesh.catNodes tetMesh.anNodes],…
ones(tetMesh.nDirichlet,1),tetMesh.nDirichlet,tetMesh.nNodes);
[B,~] = pdenullorth(H);
% form the reduced matrix algebra problem
Kc = B’*FEM.K*B;
Fc = B’*(FEM.F-(FEM.K*Vd));
%compute the FE solution
u = B*(KcFc) + Vd;
This works fine – I can now plot the nodal solutions (the electric potential) over the problem domain:
Now, I would like to compute the derived quantities associated with normal electrostatics solutions – namely the Electric Field, which is simply the gradient of the Electric Potential Field.
Ordinarily result = solvepde(model) would compute this quantity automatically – but I computed my solution simply using . Is there a way to perform this using the FE mesh that the problem was solved on? I have a hunch that the shape functions and their derivatives embedded in the FE mesh will produce a more accurate gradient field than if I project the solution to a uniform grid, compute a gradient, and project back. For this reason I would like to avoid using scatteredInterpolant(). partial differential equations, gradient, electric field, finite element mesh, post-process, pde modeling MATLAB Answers — New Questions
Challenges in Uploading Files Over 2GB via HTTP Protocol in IIS Web Server
Recently, I had worked on a case where customer`s ask was to upload a file larger than 2gb into the IIS hosted web application.
In my case, customer had a normal Asp.net Framework 4.6 application and when they were trying to perform the upload operation of more than 2gb zip file(the extension did not matter…we tested it with .7z and .zip both and it had failed) and somehow on the browser page, we got a Bad request and when I checked the HAR file, i could see a 400 status code.
Next, I checked the application configurations…like:
system.webServer/security/requestFiltering/requestLimits/maxAllowedContentLength
system.web/httpRuntime/executionTimeout
system.web/httpRuntime/maxRequestLength
but these values were already configured for the higher number.
Articles for references:
MaxAllowedContentLength
Request Filtering
Also, we checked the App Pool Managed Pipeline Mode, and the interesting part was that we got different error for both the modes.
I was able to recreate this scenario in my lab machine for a sample asp.net framework web application and I had captured FREB logs (Failed request tracing logs) for the site for both the scenarios these were the differences I noticed:
In both the cases, I was able to upload only 2gb file size.
In Integrated Mode, when I tried to upload a 4gb file, I could see this error under the FREB log: ”
400 Bad request-Asp.net detected invalid characters in the Url”. So, I tried to increase the value of “maxAllowedContentLength” to the maximum supported value, which is 4gb but still it failed, and it seems that webengine code (webengine4!MgdGetRequestBasics) doesn’t support more than 2 GB content-length.
Next, in Classic Mode, when I tried to upload a 4gb file, i got this error under FREB log : “413.1 -request entity too large error”. So, in this case also, I increased the value of “maxAllowedContentLength” to 4gb, but it failed. We did not see any error under FREB log…it was also 200 status code but the file did not upload.
So, the conclusion is that either you keep the application pipeline mode as Integrated or Classic, you would only be able to upload a 2gb file for your web application hosted on IIS.
If you would like to perform a larger file upload operation, HTTP protocol isn’t the right one. You need to switch to webDav feature, or use FTP protocol that is meant to perform the file upload/download operation without any size limit or you can keep using Http protocol but you need to send the data packets as small chunks from client to the server side and then on the server side code, you need to bundle all the chunked packets together for the file upload operation.
Also, note that even if you think moving to latest windows server would fix it…that’s not going to help here. This behavior will be same for all the supported IIS across all the versions of supported windows server.
Hope this article helps everyone.
Microsoft Tech Community – Latest Blogs –Read More
Port 80 Occupied???
Recently I was working on an issue, where customer was not able to browse the Default website hosted on IIS over port 80 and the error was:
Initial thought was, definitely the port is blocked…but who was blocking the port was the biggest question???
So, I started the investigation with basic isolation steps:
Created a different test site with binding port 80 (by stopping the Default website as we cannot run two sites with the same port together). But the test site also failed with the same error.
Created a different IP: Port 80 binding and it worked. Tested it on the default website and there also it worked.
Checked on the customers machine if they had enabled any sort of proxy or firewall that might be blocking the port 80 particularly but that wasn’t the case for the customer.
I also ran the Port Query tool and there also I got the error that Port is being used: Port Query Tool
Next, ran the below command to see what all services are listening on port 80 and nothing was useful:
netstat -ano |findstr 80
After doing some research, I found out that we have a service on the windows server called as “Sync Share service ” that also by default uses the same port number 80 as IIS Default Web Site.
Now, what is a Sync Share service???
Work Folders is available in Windows Server 2012 R2 Essentials as part of the File and Storage Services role. Work Folders uses the IIS Hostable Web Core feature and all management is performed via the Work Folders canvas in Server Manager as well as via Windows PowerShell cmdlets. Windows Server Essentials is managed via its dashboard and the IIS Management UX. Both products assume exclusive access of the SSL port (443) and HTTP port (80). This is the default configuration for both products.
In my case, the customer had the service enabled but was not using it, so we stopped the service from the Services.msc window and continued using the Port 80 for the IIS hosted web applications.
Now, if you would like to keep using both the services on the server, then you would need to change the port number for one of the services.
We have a very good article written about the same in this article: Port Conflict Issue in 2012 R2
Hope this article helps everyone
Microsoft Tech Community – Latest Blogs –Read More
how to replace elements in top third, middle third, and bottom third of matix
This question is soft-locked: new answers that are equivalent to already posted answers may be deleted without prior notice.
My task is the following:
Write a function called trio that takes two positive integer inputs n and m. The function returns a 3n-by-m matrix called T. The top third of T (an n by m submatrix) is all 1s, the middle third is all 2-s while the bottom third is all 3-s. See example below:
M = trio(2,4)
M =
1 1 1 1
1 1 1 1
2 2 2 2
2 2 2 2
3 3 3 3
3 3 3 3
This is the code that I wrote, but it only works for T = trio (4,3). I want my code to work for any input of n,m.
function T = trio (n, m)
T = randi (10, (3 * n) , m);
T ( 1:n , 🙂 = 1;
T ( (n+1):(end-(n-1)) , 🙂 = 2;
T ( (n+3):end, 🙂 = 3;
end
How is it possible to call out only top third, middle third, and bottom third of any matrix?
Thank you in advance.This question is soft-locked: new answers that are equivalent to already posted answers may be deleted without prior notice.
My task is the following:
Write a function called trio that takes two positive integer inputs n and m. The function returns a 3n-by-m matrix called T. The top third of T (an n by m submatrix) is all 1s, the middle third is all 2-s while the bottom third is all 3-s. See example below:
M = trio(2,4)
M =
1 1 1 1
1 1 1 1
2 2 2 2
2 2 2 2
3 3 3 3
3 3 3 3
This is the code that I wrote, but it only works for T = trio (4,3). I want my code to work for any input of n,m.
function T = trio (n, m)
T = randi (10, (3 * n) , m);
T ( 1:n , 🙂 = 1;
T ( (n+1):(end-(n-1)) , 🙂 = 2;
T ( (n+3):end, 🙂 = 3;
end
How is it possible to call out only top third, middle third, and bottom third of any matrix?
Thank you in advance. This question is soft-locked: new answers that are equivalent to already posted answers may be deleted without prior notice.
My task is the following:
Write a function called trio that takes two positive integer inputs n and m. The function returns a 3n-by-m matrix called T. The top third of T (an n by m submatrix) is all 1s, the middle third is all 2-s while the bottom third is all 3-s. See example below:
M = trio(2,4)
M =
1 1 1 1
1 1 1 1
2 2 2 2
2 2 2 2
3 3 3 3
3 3 3 3
This is the code that I wrote, but it only works for T = trio (4,3). I want my code to work for any input of n,m.
function T = trio (n, m)
T = randi (10, (3 * n) , m);
T ( 1:n , 🙂 = 1;
T ( (n+1):(end-(n-1)) , 🙂 = 2;
T ( (n+3):end, 🙂 = 3;
end
How is it possible to call out only top third, middle third, and bottom third of any matrix?
Thank you in advance. matrix manipulation, homework, soft-lock MATLAB Answers — New Questions
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Hello,
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🔖جلب الحبيب🌐jalb-alhabib.com🌐موقع الشيخ الروحاني
موقع جلب الحبيب ينشر تجارب البنات مع الشيخ الروحاني المغربي لجلب الحبيب
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