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Margarita [4]
2 years ago
7

What is the solution to the equation 3 over 1⁄2 plus P. equals 5 over 1⁄4​

Mathematics
1 answer:
Firdavs [7]2 years ago
8 0

Answer:

C hdgkddjgxjvvnCbbcBcgsjgzjvzjvhfhfh

You might be interested in
(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
An object with a starting velocity of 15 m/s accelerates at 3 m/s2. How far does the object travel within 10 seconds?
Anna007 [38]

Answer:  300 m

<u>Step-by-step explanation:</u>

distance(s)=v_it+\dfrac{1}{2}at^2 \quad \text{where}\ v_i \text{is initial velocity, a is acceleration, t is time}

s=(15)(10)+\dfrac{1}{2}(3)(10)^2

  = 150 + 150

  = 300

6 0
3 years ago
PART 1!! There’s more to it so please help me!! lesson 3.3.4 Practice: modeling: graphs of functions!
asambeis [7]

Answer:

the first cell next to 3 (time hours x) is 300

the  point is (3, 300)

the second cell is 1000

the point is (10,1000)

the values for the second table should be the same if they have the same unit rate which is 100 watts per hour

Step-by-step explanation:

If there is 100 watts perhour just multiply the  number of hours by this unit

5 0
2 years ago
PLEASE ANSWER SOON<br> Has a picture of question
earnstyle [38]
The answer is 21 which comes from 7 * 3.

Draw the four points in a complex plane. From the drawing you can see that the lenght is 4i - (-3i) = 4i + 3i = 7i, and the width is 1 - (-2) = 1 + 2 = 3.

So, the area is 7 * 3 = 21.




8 0
3 years ago
I need help im doing BODMAS and i dont really understand:
mars1129 [50]

Answer:

.6 x 7.2 + 1.4

Then, 4.32 + 1.4

=5.72

4 0
2 years ago
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