The real solutions the equation as given in the task content; x² = 225 are; +25 and -25.
<h3>What are the real solutions of the equation as given in the task content?</h3>
It follows from the task content that the real solutions of the equation as given in the task content can be determined as follows;
x² = 225
x = ± 15
Therefore, the real solutions of the equation are; +25 and -25.
Read more on real solutions of equations;
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I believe you meant: Use completing the square to determine the center and radius of the circle represented by this equation: <span>X^2+y^2-10x+6y=15
x^2 - 10x + 25 - 25 + y^2 + 6y + 9 - 9 = 15
Then:
(x-5)^2 + (y+3)^2 = 24 = 2sqrt(6)
This circle is centered at (5,-3) and has radius 2sqrt(6).</span>
-x^3 + x^2/3 + 5 - 13/x
I don't know if this is correct
The solution depends on the value of
. To make things simple, assume
. The homogeneous part of the equation is
and has characteristic equation
which admits the characteristic solution
.
For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be
. Then
So you have
This means
and so the general solution would be
This is what I get. Total will be 4187.56 with Interest 2187.56.
By using the formula:
To find amount :
A=p (1+r/n)^n×t
Where
P=2000,r=3%,n=1,t=25
So plug in and solve A=2000(1+0.03/1)^1×25
To find interest you use formula A=p+I
A=4187.56, p=2000,i= we need to find.
4187.56=2000+I
4187.56-2000=I
2187.56=i
