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3 years ago

Solve the following 3 × 3 system. Enter the coordinates of the solution below.

Mathematics

1 answer:

defon3 years ago

3 0

9514 1404 393

Answer:

(x, y, z) = (-1, 0, -3)

Step-by-step explanation:

We notice that the coefficients of z are such that elimination of the z term from the equations is made easy.

Adding equations 1 and 2:

(2x -3y -2z) +(x +3y +2z) = (4) +(-7)

3x = -3

x = -1

Adding equations 2 and 3:

(x +3y +2z) +(-4x -4y -2z) = (-7) +(10)

-3x -y = 3

Substituting for x, we get ...

(-3)(-1) -y = 3

0 = y . . . . . . . . . . . add y-3 to both sides

Then z can be found from any equation. Substituting for x and y in the second equation gives ...

-1 +2z = -7

2z = -6 . . . . . add 1

z = -3 . . . . . .divide by 2

The solution is (x, y, z) = (-1, 0, -3).

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What are the zeros of the following quadratic equation: y = 6x2 - 17x - 3

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Answer:

$\displaystyle x=-\frac{1}{6}, 3$

Step-by-step explanation:

Hi there!

$y = 6x^2 - 17x - 3$

Factor by grouping:

$y = 6x^2 - 18x+1x - 3\\y = 6x(x- 3)+1x - 3\\y = 6x(x- 3)+(x - 3)\\y = (6x+1)(x- 3)$

Let y=0. Apply the zero product property:

$0 = (6x+1)(x- 3)$

$6x+1=0\\6x=-1\\\\\displaystyle x=-\frac{1}{6}$

AND

$x-3=0\\x=3$

I hope this helps!

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Find the multiplicative inverse of 3 − 2i. Verify that your solution is corect by confirming that the product of

leonid [27]

Answer:

$\frac{3}{13} + \frac{2i}{13}$

Step-by-step explanation:

The multiplicative inverse of a complex number y is the complex number z such that (y)(z) = 1

So for this problem we need to find a number z such that

(3 - 2i) ( z ) = 1

If we take z = $\frac{1}{3-2i}$

We have that

$(3- 2i)\frac{1}{3-2i} = 1$ would be the multiplicative inverse of 3 - 2i

But remember that 2i = √-2 so we can rationalize the denominator of this complex number

$\frac{1}{3-2i } (\frac{3+2i}{3+2i } )=\frac{3+2i}{9-(4i^{2} )} =\frac{3+2i}{9-4(-1)} =\frac{3+2i}{13}$

Thus, the multiplicative inverse would be $\frac{3}{13} + \frac{2i}{13}$

The problem asks us to verify this by multiplying both numbers to see that the answer is 1:

Let's multiplicate this number by 3 - 2i to confirm:

$(3-2i)(\frac{3+2i}{13}) = \frac{9-4i^{2} }{13} =\frac{9-4(-1)}{13}= \frac{9+4}{13} = \frac{13}{13}= 1$

Thus, the number we found is indeed the multiplicative inverse of 3 - 2i

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