What is 2/z+3\z^2

If z= -1/4

1) The answer would be x=0.7 or x=

2) Your answer would be C. 2k+3=5.4

3) Your answer would be x=124.16 where the 6 is repeating or x=

4) ,

5) I am sorry, I do not understand number 5.

6) Your answer would be answer choice B. x=1.1 satisfies the equation because 7-1.5=5.5 and 1.1*5=5.5.

7) None of those equations is equal to the answer.

8) None of those equations is equal to the answer.

I hope this helps.

1) The answer would be x=0.7 or x=

2) Your answer would be C. 2k+3=5.4

3) Your answer would be x=124.16 where the 6 is repeating or x=

4) ,

5) I am sorry, I do not understand number 5.

6) Your answer would be answer choice B. x=1.1 satisfies the equation because 7-1.5=5.5 and 1.1*5=5.5.

7) None of those equations is equal to the answer.

8) None of those equations is equal to the answer.

I hope this helps.

Easiest method you can apply (Cramer's one). "Thanks me" if i've been helpful

Easiest method you can apply (Cramer's one). "Thanks me" if i've been helpful

B.  z = 2

Step-by-step explanation:

Set up the matrix like this.  I multiplied the last row by a 2 to get rid of the fractions so that's what you will see:

I used Cramer's Rule to solve for the value of y.  In order to do that you need to find the determinant of the matrix, then you need to find the determinant of the matrix after you sub the solution set into the column representing z.

Find the determinant requires that I "pick up" the first 2 columns and then drop them at the end of the matrix and do the multiplication of the majors minus the minors.  The matrix looks like this:

2    1    -1    2    1

0    2    3    0    2

-1    2    2    -1    2

The multiplication of the majors:

[(2×2×2)+(1×3×-1)+(-1×0×2)] = (8-3+0)=5

The multiplication of the minors:

[(-1×2×-1)+(2×3×2)+(2×0×1)] = (2+12+0) = 14

So the determinant of the matrix is I A I = 5 - 14 = -9

Now for the determinant of z, noted as I I.  Notice that I am replacing the laast column with the solution set this time:

2    1    -8    2    1

0    2    -6    0    2

-1    2    -8    -1    2

The multiplication of the majors:

[(2×2×-8)+(1×-6×-1)+(-8×0×2)] = (-32+6+0) = -26

The multiplication of the minors:

[(-1×2×-8)+(2×-6×2)+(-8×0×1)] = (16 - 24 + 0) = -8

So the determinant of I I = -18

Cramer's Rule is to divide I I by I A I:

So the value you need for z to solve for y is 2

B.  z = 2

Step-by-step explanation:

Set up the matrix like this.  I multiplied the last row by a 2 to get rid of the fractions so that's what you will see:

I used Cramer's Rule to solve for the value of y.  In order to do that you need to find the determinant of the matrix, then you need to find the determinant of the matrix after you sub the solution set into the column representing z.

Find the determinant requires that I "pick up" the first 2 columns and then drop them at the end of the matrix and do the multiplication of the majors minus the minors.  The matrix looks like this:

2    1    -1    2    1

0    2    3    0    2

-1    2    2    -1    2

The multiplication of the majors:

[(2×2×2)+(1×3×-1)+(-1×0×2)] = (8-3+0)=5

The multiplication of the minors:

[(-1×2×-1)+(2×3×2)+(2×0×1)] = (2+12+0) = 14

So the determinant of the matrix is I A I = 5 - 14 = -9

Now for the determinant of z, noted as I I.  Notice that I am replacing the laast column with the solution set this time:

2    1    -8    2    1

0    2    -6    0    2

-1    2    -8    -1    2

The multiplication of the majors:

[(2×2×-8)+(1×-6×-1)+(-8×0×2)] = (-32+6+0) = -26

The multiplication of the minors:

[(-1×2×-8)+(2×-6×2)+(-8×0×1)] = (16 - 24 + 0) = -8

So the determinant of I I = -18

Cramer's Rule is to divide I I by I A I:

So the value you need for z to solve for y is 2

Value of x = 1, y = 0 and z = 0

Option D is correct.

Step-by-step explanation:

3x + 4y + 6z = 3

4x + 3y + 3z = 4

5x + 6y + 7z = 5

We need to solve and find values of x, y and z.

I am using Elimination method.

Let

3x + 4y + 6z = 3    (1)

4x + 3y + 3z= 4     (2)

5x + 6y + 7z = 5   (3)

Multiply eq(1) with 4 and eq(2) with 3 and subtracting

12x+16y+24z = 12

12x + 9y +9z = 12

-     -       -        -

7y+15z = 0 (4)

Multiply eq(2) with 5 and eq(3) with 4 and subtracting

20x + 15y + 15z = 20

20x + 24y + 28z = 20

-      -         -           -

-9y -13z = 0    (5)

Multiply eq(4) with 9 and eq(5) with 7 and add both equations

63y + 135 z = 0

-63y - 91 z = 0

44z = 0 => z =0

Putting value of z in eq(5)

-9y -13z = 0

-9y -13(0) = 0

-9y = 0

y =0

Putting value of y and z in eq(1)

3x + 4y + 6z = 3

3x + 4(0) +6(0)=3

3x = 3

x =1

So, Value of x = 1, y = 0 and z = 0

Option D is correct.

Value of x = 1, y = 0 and z = 0

Option D is correct.

Step-by-step explanation:

3x + 4y + 6z = 3

4x + 3y + 3z = 4

5x + 6y + 7z = 5

We need to solve and find values of x, y and z.

I am using Elimination method.

Let

3x + 4y + 6z = 3    (1)

4x + 3y + 3z= 4     (2)

5x + 6y + 7z = 5   (3)

Multiply eq(1) with 4 and eq(2) with 3 and subtracting

12x+16y+24z = 12

12x + 9y +9z = 12

-     -       -        -

7y+15z = 0 (4)

Multiply eq(2) with 5 and eq(3) with 4 and subtracting

20x + 15y + 15z = 20

20x + 24y + 28z = 20

-      -         -           -

-9y -13z = 0    (5)

Multiply eq(4) with 9 and eq(5) with 7 and add both equations

63y + 135 z = 0

-63y - 91 z = 0

44z = 0 => z =0

Putting value of z in eq(5)

-9y -13z = 0

-9y -13(0) = 0

-9y = 0

y =0

Putting value of y and z in eq(1)

3x + 4y + 6z = 3

3x + 4(0) +6(0)=3

3x = 3

x =1

So, Value of x = 1, y = 0 and z = 0

Option D is correct.