Let x represent amount invested in the higher-yielding account.
We have been given that a man puts twice as much in the lower-yielding account because it is less risky. So amount invested in the lower-yielding account would be 
.
We are also told that his annual interest is $6600 dollars. We know that annual interest for one year will be principal amount times interest rate.
, where,
I = Amount of interest,
P = Principal amount,
r = Annual interest rate in decimal form,
t = Time in years.
We are told that interest rates are 6% and 10%.
Amount of interest earned from lower-yielding account: 
.
Amount of interest earned from higher-yielding account: 
.
Let us solve for x.
Therefore, the man invested $30,000 at 10%.
Amount invested in the lower-yielding account would be 
.
Therefore, the man invested $60,000 at 6%.
I'm taking the liberty of editing your function as follows: <span>w=4r^2+6s^2, where I use " ^ " to indicate exponentiation.
The partial of w with respect to r is 8r. That with respect to s is 12s.</span>
No, 3/7 and 7/3 are not the same in the fraction line. It's because 3/7 is less than 1, while 7/3 is as much as 2 1/3.
I hope it will help you mate :)
<span>x1+x2+x3=0 if x1,x2,x3 > -5?
and the solutions are integer
x1= - 4 and x2+x3=4 we count (-4;4;0), (-4;0;4), (-4;3;1), (-4;1;3), (-4; 2;2), (-4;2;2)
x1= -3 and x2+x3=3 we count (-3; 0;3), (-3;3;0), (-3;1;2), (-3;2;1)
x1= -2 and x2+x3=2 we count (-2;0;2). (-2;2;0), (-2;1;1)
x1= -1 and x2+x3=1 we count (-1;0;1), (-1;1;0)
x1=0 and (0;0;0), (0;1;-1), (0;-1;1)
There are 42 solutions
Have fun</span>
Answer:
Subsitution
Step-by-step explanation:
We have the equations:
x=2y-1
3x+y=11
The 3 methods of solving systems of equations are:
graphing
substitution
addition/elimination
Graphing involves graphing the two equations and finding their point of intersection, which is the answer
Substitution involves isolating one of the variables in one of the equations, then substituting the resulting expression into the other equation to find both numbers
Addition/Elimination involves adding the two equations together to clear a variable, solving for the other variable, and substituting the answer to the first variable in one of the equations to find the value of the cleared variable from before
Since we have x already isolated onto one side in the first expression, we can easily substitute it as 2y-1 in the second equation.
Therefore, substitution is the best method to solve this graph.
hope this helps!
