how can i calculate in matlab
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Neha Jaiswal
le 11 Juin 2020
Commenté : Abdulrahman Farouq
le 18 Jan 2022
we borrowed $1000 at a 10% anual intrests rate. if we do not makre a payment for 2 years, and assuming there is no penalty for non payment, how much do we owe now?
4 commentaires
Walter Roberson
le 8 Oct 2021
At the MATLAB command prompt give the command
format long g
now ask to display the result again.
Réponse acceptée
Atsushi Ueno
le 11 Juin 2020
simple interest:
1000 + (1000 * 0.1) * 2
compound interest:
1000 + 1000 * 0.1 + (1000 + 1000 * 0.1) * 0.1
or
1000 * (1 + 0.1) ^ 2
as function
simple interest:
function current = simple_interest(debt, rate, period)
current = debt + (debt * rate) * period;
end
compound interest:
function current = compound_interest(debt, rate, period)
current = debt * (1 + rate) ^ period;
end
3 commentaires
Plus de réponses (4)
Roopesh R Peter
le 3 Août 2020
Modifié(e) : Rik
le 3 Août 2020
Compound the interest once a year:
debt = 1000 + 1000 * 0.1 + (1000 + 1000 * 0.1) * 0.1
1 commentaire
Rik
le 3 Août 2020
(I moved the flag content to the answer body, since it shouldn't have been a tag)
Shivani Dubey
le 17 Sep 2020
p = 1000
r = 10/100
n= 1
t = 2
S =(1.1)^2
debt = p* S
1 commentaire
ahmed hammad
le 1 Sep 2021
thanks i had a tried a different methode and it has the same result but it was rejected when i submitted it to COURSERA but when i tried to you know simplify it like you did it worked
Thanks.
Medical Imaging
le 23 Sep 2020
The problem is simple and related to finance,
Borrowed_money = 1000;
Interest_rate = 10/100;
Delay_year = 2;
debt = Borrowed_money*(1+Interest_rate)^Delay_year
Jayashri Patil
le 13 Sep 2021
clear "all"
clc
n=input('enter the number of different pipe size: ');
rho=input('enter the value of density: ');
Mu=input('enter the value of viscocity: ');
Q=input('enter the flowrate: ');
e=input('enter the epsilon: ');
g=input('enter the accleration gravity: ');
for i=1:n
D(i)=input('enter the diameter of pipe: ','s');
L=input('enter the length of pipe: ','s');
KL(i)=input('enter the value of bends: ','s');
end
for i=1:n
A(i)=pi*D(i)*D(i)/4;
v(i)=Q/A(i);
ReynoldsNumber(i)=rho*v(i)*D(i)/Mu;
if ReynoldsNumber(i) <2300
f(i)=64/ReynoldsNumber(i);
elseif ReynoldsNumber(i) >4000
c=@(f)(l/sqrt(f(i)))+(2*log10*(e/D(i)*3.7+2.51/ReynoldsNumber(i)*sqrt(f(i))));
f(i)=fzero(c,[0.01,0.05]);
end
deltap(i)=(f(i)*(L(i)/D(i))*((rho*v(i)*v(i))/2)/1000); % pressure drop in kpa
HL(i)=(f(i)*(L(i)/D(i))+KL(i)*v(i)*v(i)/2*g); % head loss in m
end
4 commentaires
Rik
le 13 Sep 2021
You already posted a very similar question. Why not post a comment there with a follow-up question?
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