The perimeter is 30 units
x=31,y=−61
Put the equations in standard form and then use matrices to solve the system of equations.
5x+4y=1,3x−6y=2
Write the equations in matrix form.
(534−6)(xy)=(12)
Left multiply the equation by the inverse matrix of (534−6).
inverse((534−6))(534−6)(xy)=inverse((534−6))(12)
The product of a matrix and its inverse is the identity matrix.
(1001)(xy)=inverse((534−6))(12)
Multiply the matrices on the left hand side of the equal sign.
(xy)=inverse((534−6))(12)
For the 2×2 matrix (acbd), the inverse matrix is (ad−bcdad−bc−cad−bc−bad−bca), so the matrix equation can be rewritten as a matrix multiplication problem.
(xy)=(5(−6)−4×3−6−5(−6)−4×33−5(−6)−4×345(−6)−4×35)(12)
Do the arithmetic.
(xy)=(71141212−425)(12)
Multiply the matrices.
(xy)=(71+212×2141−425×2)
Do the arithmetic.
(xy)=(31−61)
Extract the matrix elements x and y.
x=31,y=−61
Answer:
Step-by-step explanation:
1). If a point C is lying on a line segment RB,
Then RB = RC + CB
Since, RB = 45, CB = 3x - 12 and RC = 2x + 9
45 = (2x + 9) + (3x - 12)
45 = (2x + 3x) - 3
48 = 5x
x = 
x = 9.6
2). Since, RC = 2x + 9
RC = 2(9.6) + 9
= 19.2 + 9
= 28.2 units
3). Distance between the two points 
and 
is given by the formula,
d = 
Distance between the points A(2, -2) and D(-2, 2) will be,
= 
= 
= 5.657
≈ 5.66 units
Point d is at (-1, -2) find the average of the values to get the midpoint
Answer:
100r^4 + 400r^3 + 600r^2 + 400r + 100
Step-by-step explanation:
Expanding ( r + 1 )^4 gives :-
r^4 + 4r^3 + 6r^2 + 4r + 1
So multiplying 100 with r^4 + 4r^3 + 6r^2 + 4r + 1 gives :-
100r^4 + 400r^3 + 600r^2 + 400r + 100
