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klasskru [66]
2 years ago
11

(2*)2-3×2*+2=04m-15°(×)m+75°​

Mathematics
1 answer:
Valentin [98]2 years ago
7 0

Answer:

First, we write the augmented matrix.

⎡

⎢

⎣

1

−

1

1

2

3

−

1

3

−

2

−

9

|

8

−

2

9

⎤

⎥

⎦

Next, we perform row operations to obtain row-echelon form.

−

2

R

1

+

R

2

=

R

2

→

⎡

⎢

⎣

1

−

1

1

0

5

−

3

3

−

2

−

9

|

8

−

18

9

⎤

⎥

⎦

−

3

R

1

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

5

−

3

0

1

−

12

|

8

−

18

−

15

⎤

⎥

⎦

The easiest way to obtain a 1 in row 2 of column 1 is to interchange \displaystyle {R}_{2}R

​2

​​  and \displaystyle {R}_{3}R

​3

​​ .

Interchange

R

2

and

R

3

→

⎡

⎢

⎣

1

−

1

1

8

0

1

−

12

−

15

0

5

−

3

−

18

⎤

⎥

⎦

Then

−

5

R

2

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

1

−

12

0

0

57

|

8

−

15

57

⎤

⎥

⎦

−

1

57

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

1

−

12

0

0

1

|

8

−

15

1

⎤

⎥

⎦

The last matrix represents the equivalent system.

x

−

y

+

z

=

8

y

−

12

z

=

−

15

z

=

1

Using back-substitution, we obtain the solution as \displaystyle \left(4,-3,1\right)(4,−3,1).First, we write the augmented matrix.

⎡

⎢

⎣

1

−

1

1

2

3

−

1

3

−

2

−

9

|

8

−

2

9

⎤

⎥

⎦

Next, we perform row operations to obtain row-echelon form.

−

2

R

1

+

R

2

=

R

2

→

⎡

⎢

⎣

1

−

1

1

0

5

−

3

3

−

2

−

9

|

8

−

18

9

⎤

⎥

⎦

−

3

R

1

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

5

−

3

0

1

−

12

|

8

−

18

−

15

⎤

⎥

⎦

The easiest way to obtain a 1 in row 2 of column 1 is to interchange \displaystyle {R}_{2}R

​2

​​  and \displaystyle {R}_{3}R

​3

​​ .

Interchange

R

2

and

R

3

→

⎡

⎢

⎣

1

−

1

1

8

0

1

−

12

−

15

0

5

−

3

−

18

⎤

⎥

⎦

Then

−

5

R

2

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

1

−

12

0

0

57

|

8

−

15

57

⎤

⎥

⎦

−

1

57

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

1

−

12

0

0

1

|

8

−

15

1

⎤

⎥

⎦

The last matrix represents the equivalent system.

x

−

y

+

z

=

8

y

−

12

z

=

−

15

z=1

Using back-substitution, we obtain the solution as \displaystyle \left(4,-3,1\right)(4,−3,1).

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Can someone help me with this<br> 2x+4y+3z=6<br> x-2y+z=-5<br> -x-3y-2z=-7<br> solve for x,y,z
Flauer [41]

Answer:

x = -7 2/3, y = 1 1/3 and z = 5 1/3.

Step-by-step explanation:

2x+4y+3z=6    .....  1

x-2y+z=-5        ...... 2

-x-3y-2z=-7      .......3

Add equations 2 and 3 to eliminate x:

-5y - z = -12   .....4

Multiply equation 2 by - 2:

-2x + 4y  - 2z = 10

Add this to equation 1:

8y + z = 16   ........ 5

Now add equation 4 to equation 5:

3y = 4

y = 4/3 = 1 1/3.

Now find z by substituting for y in equation 4:

-5(4/3) - z = -12

z = 12 - 20/3

z = 36/3 - 20/3 = 16/3 = 5 1/3.

Finally, we find x by substituting for y and z in equation 1:

2x + 4*4/3 + 3*16/3 = 6

2x = 6 - 16/3 - 16

2x =  18/3 - 16/3 - 48/3 = -46/3

x = 23/3 =  7 2/3.

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Answer:

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liq [111]

Check the picture below.

since the diameter of the cone is 6", then its radius is half that or 3", so getting the volume of only the cone, not the top.

1)

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2)

now, the top of it, as you notice in the picture, is a semicircle, whose radius is the same as the cone's, 3.

\bf \textit{volume of a sphere}\\\\ V=\cfrac{4\pi r^3}{3}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=3 \end{cases}\implies V=\cfrac{4\pi (3)^3}{3}\implies V=36\pi \\\\\\ \stackrel{\textit{half of that for a semisphere}}{V=18\pi }\implies V\approx 56.55

3)

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4)

pretty much the same thing, we get the volume of the cone and its top, add them up.

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alexandr1967 [171]
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This means that each of the shares is valid the whole sum divided by the number of shares, that is:
\frac{100000}{2500} = \frac{1000}{25} =40
(first i divided both denominator and numerator by 100)
So each value is worth 40 dolars.
4 0
3 years ago
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