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Leviafan [203]
2 years ago
12

PLEASE ANSWER IXL QUESTION,WILL GIVE BRAINLITIS!!!!! its attached

Mathematics
1 answer:
SpyIntel [72]2 years ago
6 0

Answer:

y=-x/3+7/3

Step-by-step explanation:

Use linear function: y=mx+b

m= -1/3

Substitue

m= -1/3,(8,-5)

5= -1/3 * (-8) +b

Solve the equation

5= -1/3 *(-8) +b

Rearrange unknown terms to the left side of the equation

-b= -1/3 * (-8) -5

Reduce the greatest common factor on both sides of the equation

b= 1/3 * (-8) +5

Remove the parentheses

b= -1/3 *8+5

Write as a single fraction

b= -8/3+5

Convert the expression into a fraction

b=-8/3+5/1

Expand the fraction to get the least common denominator

b= -8/3+5*3/1*3

Calculate product or quotient

b=-8/3+15/3

Write the numerators over the common denominator

b=-8+15/3

Calculate the sum or difference

b=7/3

Subsitute

m= -1/3,b=7/3

y= -1/3x+7/3

Rewrite the equation

y= -1/3x+7/3

Find the required form

y=-x/3+7/3

Hope this helps!!

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Determine whether the relationship between the circumfrance of a circle and its diameter is a direct variation. If so, identify
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The relationship between the circumference of a circle and its diameter represent  a direct variation and the constant of proportionality is equal to the constant \pi

Step-by-step explanation:

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3 years ago
Perform the indicated row operations, then write the new matrix.
Studentka2010 [4]

The matrix is not properly formatted.

However, I'm able to rearrange the question as:

\left[\begin{array}{ccc}1&1&1|-1\\-2&3&5|3\\3&2&4|1\end{array}\right]

Operations:

2R_1 + R_2 ->R_2

-3R_1 +R_3 ->R_3

Please note that the above may not reflect the original question. However, you should be able to implement my steps in your question.

Answer:

\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]

Step-by-step explanation:

The first operation:

2R_1 + R_2 ->R_2

This means that the new second row (R2) is derived by:

Multiplying the first row (R1) by 2; add this to the second row

The row 1 elements are:

\left[\begin{array}{ccc}1&1&1|-1\end{array}\right]

Multiply by 2

2 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}2&2&2|-2\end{array}\right]

Add to row 2 elements are: \left[\begin{array}{ccc}-2&3&5|3\end{array}\right]

\left[\begin{array}{ccc}2&2&2|-2\end{array}\right] + \left[\begin{array}{ccc}-2&3&5|3\end{array}\right]

\left[\begin{array}{ccc}0&5&7|1\end{array}\right]

The second operation:

-3R_1 +R_3 ->R_3

This means that the new third row (R3) is derived by:

Multiplying the first row (R1) by -3; add this to the third row

The row 1 elements are:

\left[\begin{array}{ccc}1&1&1|-1\end{array}\right]

Multiply by -3

-3 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right]

Add to row 2 elements are: \left[\begin{array}{ccc}3&2&4|1\end{array}\right]

\left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right] + \left[\begin{array}{ccc}3&2&4|1\end{array}\right]

\left[\begin{array}{ccc}0&-1&1|4\end{array}\right]

Hence, the new matrix is:

\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]

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3 years ago
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