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Katen [24]
4 years ago
10

How to multiply a matrices 3x3 by 3x2

Mathematics
1 answer:
lesya692 [45]4 years ago
5 0
The top row of matrix A (1, 2, 1) is multiplied with the first column of matrix B (1,0,-1) and the result is 1x1 + 2x0 + 1x -1 = 0 this is row 1 column 1 of the resultant matrix
The top row of matrix A (1,2,1) is multiplied with the second column of matrix B (-1, -1, 1) and the result is 1 x-1 + 2 x -1 + 1 x 1 = -2 , this is row 1 column 2 of the resultant matrix
Repeat with the second row of matrix A (-1,-1.-2) x (1,0,-1) = 1 this is row 2 column 1 of the resultant matrix, multiply the second row of A (-1,-1,-2) x (-1,-1,1) = 0, this is row 2 column 2 of the resultant
Repeat with the third row of matrix A( -1,1,-2) x (1,0, -1) = 1, this is row 3 column 1 of the resultant
the third row of A (-1,1,-2) x( -1,-1,1) = -2, this is row 3 column 2 of the resultant matrix

Matrix AB ( 0,-2/1,0/1,-2) 
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Find a polynomial of degree 3 with a real coefficient that satisfies the given condition. Zeros: -2, 1, 0; f(2)=32
xz_007 [3.2K]
Its factors would be
(x+2)*(x-1)*(x+0)

x^2 +x -2

x^3 + 0 + x^2 + 0 -2x +0

Equation: x^3 + x^2 -2x
f(2) = 8 + 4 -4

2x^3  + 2x^2  -4x +0
f(2) = 16 + 8 -8

3x^3  + 3x^2  -6x +0
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So, the equation is:
4x^3  + 4x^2  -8x = 0




8 0
3 years ago
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Find the a15 of the sequence 7/8, 1 / 2, 1 / 8, -1 / 4​
Dennis_Churaev [7]

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Find the equation of the circle with center (4,5) which passes through the y-intercept of the line 5x-2y+6=0​
Elena-2011 [213]

Answer:

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Step-by-step explanation:

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We know the Circle has the formula (x-4)^2 + (y-5)^2 = ?^2 from the question but with the intercept, we can find the entire equation as the y-intercept is (0,3) so we can substitute it into the equation to find the full equations so:

(0-4)^2 + (3-5)^2 = ?^2

This simplifies to:

(-4)^2 +(-2)^2 = ?^2

16 + 4 = 20 = ?^2

The answer is 20 so the equation of the circle is (x-4)^2 + (y-5)^2 = 20

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