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tensa zangetsu [6.8K]
4 years ago
13

Y=3.6x. Tell whether this equation represents a direct variation. If so, identify the constant of variation

Mathematics
1 answer:
ehidna [41]4 years ago
6 0
Yes it is
constant =3.6
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Which ratio is equivalent to five nine
salantis [7]

Answer:

30:54

Step-by-step explanation:

5:9 x 6 = 30:54

3 0
3 years ago
Find the distance of each the following lines from the origin.
meriva

Answer:

Step-by-step explanation:

3x+y−7=0 and x+2y+9=0

write in the form y=mx+c

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7 0
3 years ago
Explain how 49. 567 is done in fraction form
pav-90 [236]
<span>= 49567/1000

</span>rite the decimal number as a fraction
(over 1)
49.567 = 49.567 / 1

Multiplying by 1 to eliminate 3 decimal places
we multiply top and bottom by 3 10's

Numerator (N)
N = 49.567 × 10 × 10 × 10 = 49567
Denominator (D)
D = 1 × 10 × 10 × 10 = 1000

N / D = 49567 / 1000

Simplifying our fraction

<span>= 49567/1000</span>
7 0
3 years ago
A2 = [1 2 3; 4 5 6; 7 8 9; 3 2 4; 6 5 4; 9 8 7]
34kurt

Answer:

Remember, a basis for the row space of a matrix A is the set of rows different of zero of the echelon form of A.

We need to find the echelon form of the matrix augmented matrix of the system A2x=b2

B=\left[\begin{array}{cccc}1&2&3&1\\4&5&6&1\\7&8&9&1\\3&2&4&1\\6&5&4&1\\9&8&7&1\end{array}\right]

We apply row operations:

1.

  • To row 2 we subtract row 1, 4 times.
  • To row 3 we subtract row 1, 7 times.
  • To row 4 we subtract row 1, 3 times.
  • To row 5 we subtract row 1, 6 times.
  • To row 6 we subtract row 1, 9 times.

We obtain the matrix

\left[\begin{array}{cccc}1&2&3&1\\0&-3&-6&-3\\0&-6&-12&-6\\0&-4&-5&-2\\0&-7&-14&-5\\0&-10&-20&-8\end{array}\right]

2.

  • We subtract row two twice to row three of the previous matrix.
  • we subtract 4/3 from row two to row 4.
  • we subtract 7/3 from row two to row 5.
  • we subtract 10/3 from row two to row 6.

We obtain the matrix

\left[\begin{array}{cccc}1&2&3&1\\0&-3&-6&-3\\0&0&0&0\\0&0&3&2\\0&0&0&2\\0&0&0&2\end{array}\right]

3.

we exchange rows three and four of the previous matrix and obtain the echelon form of the augmented matrix.

\left[\begin{array}{cccc}1&2&3&1\\0&-3&-6&-3\\0&0&3&2\\0&0&0&0\\0&0&0&2\\0&0&0&2\end{array}\right]

Since the only nonzero rows of the augmented matrix of the coefficient matrix are the first three, then the set

\{\left[\begin{array}{c}1\\2\\3\end{array}\right],\left[\begin{array}{c}0\\-3\\-6\end{array}\right],\left[\begin{array}{c}0\\0\\3\end{array}\right] \}

is a basis for Row (A2)

Now, observe that the last two rows of the echelon form of the augmented matrix have the last coordinate different of zero. Then, the system is inconsistent. This means that the system has no solutions.

4 0
3 years ago
Need help! Zoom if necessary
arlik [135]
The first one B,D,C,A
3 0
3 years ago
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