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Sladkaya [172]
3 years ago
7

1/3 21 - 1/2 12 can somebody please help me!

Mathematics
2 answers:
Alexxandr [17]3 years ago
7 0

Answer:

1

Step-by-step explanation:

Simplify the following:

7 - 1/2×12

(-1)/2×12 = (-12)/2:

7 + (-12)/2

12/2 = (2×6)/2 = 6:

7 - 6

21/3 = (3×7)/3 = 7:

7 - 6

7 - 6 = 1:

Answer:  1

aleksklad [387]3 years ago
3 0

Answer:

The answer is 8 and five-sixths


Step-by-step explanation:


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Doreen schmidt is a chemist. She needs to prepare 36 ounces of a 10% shydrochloric acid solution. Find the amount of 18% solutio
emmainna [20.7K]
Total = 36oz
Amount of 18% solution = x
Amount of 9% solution = 36 - x

Using the fact that (amount of solution) * (percent HCl) = amount HCl we can set up an expression for the amount HCl in a mixture of 18% and 9% solutions with a total volume of 36oz.

Amount of HCl = x*0.18 + (36 - x)*0.09

If we divide this by the amount of solution we get percent HCl, and we have a target of 10% HCl. Now we have an equation to solve,

[x*0.18 + (36 - x)*0.09]/36 = 0.1

x*0.09 + 3.24 = 3.6

.09x = 0.36
x = 4

Therefore,
4 ounces of 18% HCl solution and 32 ounces of 9% HCl solution

Check:
[(4*.18) + (.09*32)]/36 = 0.1?
0.1 = 0.1   Yes.

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(2*)2-3×2*+2=0<br>4m-15°(×)m+75°​
Valentin [98]

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).

7 0
2 years ago
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