Answer: 4a + b - 5
Step-by-step explanation:
We can only simplify the like terms of 3a and a, which add to 4a and we keep the rest the same.
Answer:
:
1.
centre(h,k)=(-13,9)
radius (r)=6
we have
equation of the circle is
(x-h)²+(y-k)²=r²
(x+13)²+(y-9)²=6²
(x+13)²+(y-9)²=36you can write this equation too
x²+26x+169+y²-18y+81=36
x²+y²+26x-18y+169+81-36=0
x²+y²+26x-18y +214=0
is a required equation of the circle.
2.
centre(h,k)=(1,-1)
radius (r)=11
we have
equation of the circle is
(x-h)²+(y-k)²=r²
(x-1)²+(y+1)²=11²
(x-1)²+(y+1)²=121
you can write this equation too
x²-2x+1+y²+2y+1=121
x²+y²-2x+2y=121-2
x²+y²-2x+2y=119
is a required equation of the circle.
3.
centre (h,k)=(3,1)
radius (r)=4units.
we have
equation of the circle is
(x-h)²+(y-k)²=r²
(x-3)²+(y-1)²=4²
(x-3)²+(y-1)²=16 is a required equation.
4.
centre(h,k)=(4,-2)
radius (r)=3
we have
equation of the circle is
(x-h)²+(y-k)²=r²
(x-4)²+(y+2)²=3²
(x-4)²+(y+2)²=9
you can write this equation
at maximum. dt/dh=0
if you differentiate you will get
-32t=-100
t=3.125
By algebraic handling we find that the unique solution of the system of linear equations is: (x, y, z) = (- 2, 4, 3). (Correct choice: A)
<h3>What is the nature of a system of linear equations?</h3>
If the system of equations has no solutions, then the determinant of the dependent coefficients of the system of linear equations must be zero. Let see:
This determinant can be determined by Sarrus' rule:
D = (1) · (- 1) · (7) + 4 · (- 1) · 1 + (- 1) · 1 · (- 8) - (- 1) · (- 1) · 1 + 4 · 1 · 7 - 1 · (- 1) · (- 8)
D = - 7 - 4 + 8 - 1 + 28 - 8
D = 16
The system of linear equations have at least one solution. This system has only one solution any of the three equations is not a function of the other two. By algebraic handling we find that the unique solution of the system is: (x, y, z) = (- 2, 4, 3).
To learn more on systems of linear equations: brainly.com/question/19549073
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