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OleMash [197]
3 years ago
5

Graph ​ y=−5/6x−4 ​. Use the line tool and select two points on the line.

Mathematics
2 answers:
Aleksandr [31]3 years ago
6 0

Here is a image of the graph

creativ13 [48]3 years ago
4 0

y =  \frac{ - 5}{6x - 4}
so I think I fill 0 in for Y and 0 in for X so:
0 =  \frac{ - 5}{6x - 4}
you multiply 6x-4 to the 0
0 = -5

no solution

then you do the same but with X

y =  \frac{ - 5}{6(0) - 4}
y =  \frac{ - 5}{0 - 4}
y =  \frac{ - 5}{ - 4}
negatives cancel out
y =  \frac{5}{4}


I would check this! I am not totally sure!!
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Kindly check explanation

Step-by-step explanation:

When performing addition and subtraction of decimals, it is important to arrange the numbers being added or subtracted such that the decimal points are in line. This is particularly important so that the place value of the numbers are in accord.

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4 0
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UNO [17]
<h2><u>ABSOLUTE VALUE</u></h2>

The absolute value of a number is the distance from 0 to that number. The distance is positive, hence, the absolute value is always a positive number.

<h3>Exercise</h3>

Replace the value of x:

| x - 8 |

| -4 - 8 |

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3 years ago
Read 2 more answers
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inn [45]

Answer:

Step-by-step explanation:

2005 AMC 8 Problems/Problem 20

Problem

Alice and Bob play a game involving a circle whose circumference is divided by 12 equally-spaced points. The points are numbered clockwise, from 1 to 12. Both start on point 12. Alice moves clockwise and Bob, counterclockwise. In a turn of the game, Alice moves 5 points clockwise and Bob moves 9 points counterclockwise. The game ends when they stop on the same point. How many turns will this take?

$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 24$

Solution

Alice moves $5k$ steps and Bob moves $9k$ steps, where $k$ is the turn they are on. Alice and Bob coincide when the number of steps they move collectively, $14k$, is a multiple of $12$. Since this number must be a multiple of $12$, as stated in the previous sentence, $14$ has a factor $2$, $k$ must have a factor of $6$. The smallest number of turns that is a multiple of $6$ is $\boxed{\textbf{(A)}\ 6}$.

See Also

2005 AMC 8 (Problems • Answer Key • Resources)

Preceded by

Problem 19 Followed by

Problem 21

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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