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Usimov [2.4K]
3 years ago
14

How would 6 3/6 be classified

Mathematics
2 answers:
gulaghasi [49]3 years ago
4 0
If it's math then 6 3/6 would be a mixed number
Zinaida [17]3 years ago
4 0
Rational and real,. b is your answer for sure
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Just simplify?
Than your answer should be m=- \frac{18}{17}
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What is the quotient 12/4 as a fraction
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The answer is 4/12
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Which value of m will create a system of parallel lines with no solution? y=mx-6 8x-4y=12 A coordinate grid with one line labele
nydimaria [60]

Answer:

A system of parallel lines will be created where the two lines will never meet and have no common solution at a value of m = 2

Step-by-step explanation:

The equation of the given line is 8·x - 4·y = 12

Which gives;

8·x- 12= 4·y

y = 2·x - 3

Given that the line passes through the points (0, -3) and (1, -1), we have;

Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}

When (x₁, y₁) = (0. -3) and (x₂, y₂) = (1, -1), we have;

Slope, \, m =\dfrac{(-1)-(-3)}{1-(0)} = 2

y - (-3) = 2×(x - 0)

y = 2·x - 3 which is the equation of the given line

For the lines 8·x - 4·y = 12, which is the sane as y = 2·x - 3 and the line y = m·x  - 6 to have no solution, the slope of the two lines should be equal that is m = 2

Given that the line passes through the point (1.5, 0), we have;

y - 0 = 2×(x - 1.5)  

y = 2·x - 3...................(1)

For the equation, y = m·x  - 6, when m = 2, we have;

y = 2·x  - 6..................(2)

Solving equations (1) and (2) gives;

2·x - 3 = 2·x  - 6, which gives;

2·x - 2·x=  - 3 - 6

0 = 9

Therefore, a system of parallel lines will be created where the two lines will never meet and have no common solution at a value of m = 2.

4 0
3 years ago
Polygon A is similar to Polygon B. Find the perimeter of Polygon B.
marissa [1.9K]
\bf \textit{ratios relations for similar figures}
\\\\\\
\begin{array}{cccllll}
side&area&volume\\
\textendash\textendash\textendash\textendash\textendash\textendash&\textendash\textendash\textendash\textendash\textendash\textendash&\textendash\textendash\textendash\textendash\textendash\textendash\\
\frac{s}{s}=\frac{s}{s}&\frac{s^2}{s^2}=\frac{area}{area}&\frac{s^3}{s^3}=\frac{volume}{volume}
\end{array}\\\\
-----------------------------\\\\


\bf thus
\\\\\\
\begin{array}{llll}
\cfrac{\textit{polygon A}}{\textit{polygon B}}\qquad &\cfrac{24^2}{15^2}=\cfrac{128}{x}
\end{array}

solve for "x"
3 0
3 years ago
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