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Jse the graph or table to find the equation that represents the relationship.
CLEAR
CHECK
NEXT
>
回Referen
calculator

(x)
formulas
glossary
Representation
Equation

x

y

2
1

1
-2

3
4

DRAG AND DROP
AN ITEM HERE

x

y

(1)/(4)
-1

(1)/(8)

(1)/(2)

0
2

DRAG AND DROP
AN ITEM HERE
DRAG AND DROP
AN ITEM HERE
DRAG AND DROP
AN ITEM HERE

Jse the graph or table to find the equation that represents the relationship. $\newline$ CLEAR $\newline$ CHECK $\newline$ NEXT $>$ $\newline$ 回Referen $\newline$ calculator $\newline$ $(x)$ $\newline$ formulas $\newline$ glossary $\newline$ Representation $\newline$ Equation $\newline$ \begin{tabular}{|c|c|} $\newline$ \hline $x$ & $y$ \\ $\newline$ \hline $2$ & $1$ \\ $\newline$ \hline $1$ & $-2$ \\ $\newline$ \hline $3$ & $4$ \\ $\newline$ \hline $\newline$ \end{tabular} $\newline$ DRAG AND DROP $\newline$ AN ITEM HERE $\newline$ \begin{tabular}{|c|c|} $\newline$ \hline $x$ & $y$ \\ $\newline$ \hline $\frac{1}{4}$ & $-1$ \\ $\newline$ \hline $\frac{1}{8}$ & $\frac{1}{2}$ \\ $\newline$ \hline $0$ & $2$ \\ $\newline$ \hline $\newline$ \end{tabular} $\newline$ DRAG AND DROP $\newline$ AN ITEM HERE $\newline$ DRAG AND DROP $\newline$ AN ITEM HERE $\newline$ DRAG AND DROP $\newline$ AN ITEM HERE

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Math Problems

Grade 6

Estimate quotients when dividing mixed numbers

Full solution

Q. Jse the graph or table to find the equation that represents the relationship.

\newline

CLEAR

\newline

CHECK

\newline

>

\newline

回Referen

\newline

calculator

\newline

(x)

\newline

formulas

\newline

glossary

\newline

Representation

\newline

Equation

\newline

\begin{tabular}{|c|c|}

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\hline

x

y

\newline

\hline

2

1

\newline

\hline

1

-2

\newline

\hline

3

4

\newline

\hline

\newline

\end{tabular}

\newline

DRAG AND DROP

\newline

AN ITEM HERE

\newline

\begin{tabular}{|c|c|}

\newline

\hline

x

y

\newline

\hline

\frac{1}{4}

-1

\newline

\hline

\frac{1}{8}

\frac{1}{2}

\newline

\hline

0

2

\newline

\hline

\newline

\end{tabular}

\newline

DRAG AND DROP

\newline

AN ITEM HERE

\newline

DRAG AND DROP

\newline

AN ITEM HERE

\newline

DRAG AND DROP

\newline

AN ITEM HERE

Analyze Data: First, let's look at the pairs of $x$ and $y$ values to see if there's a pattern.
Identify Pattern: We have the points $(2, 1)$ , $(1, -2)$ , $(3, 4)$ , $(\frac{1}{4}, -1)$ , $(\frac{1}{8}, \frac{1}{2})$ , and $(0, 2)$ .
Calculate Differences: Let's try to find the difference between the $y$ -values when the $x$ -value increases by $1$ . From $(1, -2)$ to $(2, 1)$ , the $y$ -value increases by $3$ .
Plot Points: But when we look at the other points, like $(\frac{1}{4}, -1)$ to $(\frac{1}{8}, \frac{1}{2})$ , the pattern doesn't hold up. So it's not a simple linear relationship.
Consider Quadratic Equation: Let's plot the points on a graph to see if we can visualize a pattern.
Find Constant Term: After plotting, it seems like the points might fit a quadratic equation, since they don't line up in a straight line.
Create System of Equations: To find a quadratic equation, we need to find $a$ , $b$ , and $c$ for the equation $y = ax^2 + bx + c$ .
Substitute Points: Using the point $(0, 2)$ , we can immediately find that $c = 2$ because when $x$ is $0$ , $y$ is $2$ .
Solve for Coefficients: Now we have $y = ax^2 + bx + 2$ . We need $2$ more points to create a system of equations to solve for $a$ and $b$ .
Finalize Equation: Using the point $(1, -2)$ , we substitute into the equation to get $-2 = a(1)^2 + b(1) + 2$ .
Finalize Equation: Using the point $(1, -2)$ , we substitute into the equation to get $-2 = a(1)^2 + b(1) + 2$ . This simplifies to $-2 = a + b + 2$ . Subtracting $2$ from both sides, we get $a + b = -4$ .
Finalize Equation: Using the point $(1, -2)$ , we substitute into the equation to get $-2 = a(1)^2 + b(1) + 2$ . This simplifies to $-2 = a + b + 2$ . Subtracting $2$ from both sides, we get $a + b = -4$ . Using the point $(2, 1)$ , we substitute into the equation to get $1 = a(2)^2 + b(2) + 2$ .
Finalize Equation: Using the point $(1, -2)$ , we substitute into the equation to get $-2 = a(1)^2 + b(1) + 2$ . This simplifies to $-2 = a + b + 2$ . Subtracting $2$ from both sides, we get $a + b = -4$ . Using the point $(2, 1)$ , we substitute into the equation to get $1 = a(2)^2 + b(2) + 2$ . This simplifies to $1 = 4a + 2b + 2$ . Subtracting $2$ from both sides, we get $4a + 2b = -1$ .
Finalize Equation: Using the point $(1, -2)$ , we substitute into the equation to get $-2 = a(1)^2 + b(1) + 2$ . This simplifies to $-2 = a + b + 2$ . Subtracting $2$ from both sides, we get $a + b = -4$ . Using the point $(2, 1)$ , we substitute into the equation to get $1 = a(2)^2 + b(2) + 2$ . This simplifies to $1 = 4a + 2b + 2$ . Subtracting $2$ from both sides, we get $4a + 2b = -1$ . Now we have the system of equations: $a + b = -4$ and $4a + 2b = -1$ .
Finalize Equation: Using the point $(1, -2)$ , we substitute into the equation to get $-2 = a(1)^2 + b(1) + 2$ . This simplifies to $-2 = a + b + 2$ . Subtracting $2$ from both sides, we get $a + b = -4$ . Using the point $(2, 1)$ , we substitute into the equation to get $1 = a(2)^2 + b(2) + 2$ . This simplifies to $1 = 4a + 2b + 2$ . Subtracting $2$ from both sides, we get $4a + 2b = -1$ . Now we have the system of equations: $a + b = -4$ and $4a + 2b = -1$ . We can multiply the first equation by $2$ to get $-2 = a(1)^2 + b(1) + 2$ $3$ and then subtract it from the second equation.
Finalize Equation: Using the point $(1, -2)$ , we substitute into the equation to get $-2 = a(1)^2 + b(1) + 2$ . This simplifies to $-2 = a + b + 2$ . Subtracting $2$ from both sides, we get $a + b = -4$ . Using the point $(2, 1)$ , we substitute into the equation to get $1 = a(2)^2 + b(2) + 2$ . This simplifies to $1 = 4a + 2b + 2$ . Subtracting $2$ from both sides, we get $4a + 2b = -1$ . Now we have the system of equations: $a + b = -4$ and $4a + 2b = -1$ . We can multiply the first equation by $2$ to get $-2 = a(1)^2 + b(1) + 2$ $3$ and then subtract it from the second equation. Subtracting we get $-2 = a(1)^2 + b(1) + 2$ $4$ , which simplifies to $-2 = a(1)^2 + b(1) + 2$ $5$ .
Finalize Equation: Using the point $(1, -2)$ , we substitute into the equation to get $-2 = a(1)^2 + b(1) + 2$ . This simplifies to $-2 = a + b + 2$ . Subtracting $2$ from both sides, we get $a + b = -4$ . Using the point $(2, 1)$ , we substitute into the equation to get $1 = a(2)^2 + b(2) + 2$ . This simplifies to $1 = 4a + 2b + 2$ . Subtracting $2$ from both sides, we get $4a + 2b = -1$ . Now we have the system of equations: $a + b = -4$ and $4a + 2b = -1$ . We can multiply the first equation by $2$ to get $-2 = a(1)^2 + b(1) + 2$ $3$ and then subtract it from the second equation. Subtracting we get $-2 = a(1)^2 + b(1) + 2$ $4$ , which simplifies to $-2 = a(1)^2 + b(1) + 2$ $5$ . Dividing both sides by $2$ , we find that $-2 = a(1)^2 + b(1) + 2$ $7$ .
Finalize Equation: Using the point $(1, -2)$ , we substitute into the equation to get $-2 = a(1)^2 + b(1) + 2$ . This simplifies to $-2 = a + b + 2$ . Subtracting $2$ from both sides, we get $a + b = -4$ . Using the point $(2, 1)$ , we substitute into the equation to get $1 = a(2)^2 + b(2) + 2$ . This simplifies to $1 = 4a + 2b + 2$ . Subtracting $2$ from both sides, we get $4a + 2b = -1$ . Now we have the system of equations: $a + b = -4$ and $4a + 2b = -1$ . We can multiply the first equation by $2$ to get $-2 = a(1)^2 + b(1) + 2$ $3$ and then subtract it from the second equation. Subtracting we get $-2 = a(1)^2 + b(1) + 2$ $4$ , which simplifies to $-2 = a(1)^2 + b(1) + 2$ $5$ . Dividing both sides by $2$ , we find that $-2 = a(1)^2 + b(1) + 2$ $7$ . Substituting $-2 = a(1)^2 + b(1) + 2$ $8$ back into the first equation, we get $-2 = a(1)^2 + b(1) + 2$ $9$ .
Finalize Equation: Using the point $(1, -2)$ , we substitute into the equation to get $-2 = a(1)^2 + b(1) + 2$ . This simplifies to $-2 = a + b + 2$ . Subtracting $2$ from both sides, we get $a + b = -4$ . Using the point $(2, 1)$ , we substitute into the equation to get $1 = a(2)^2 + b(2) + 2$ . This simplifies to $1 = 4a + 2b + 2$ . Subtracting $2$ from both sides, we get $4a + 2b = -1$ . Now we have the system of equations: $a + b = -4$ and $4a + 2b = -1$ . We can multiply the first equation by $2$ to get $-2 = a(1)^2 + b(1) + 2$ $3$ and then subtract it from the second equation. Subtracting we get $-2 = a(1)^2 + b(1) + 2$ $4$ , which simplifies to $-2 = a(1)^2 + b(1) + 2$ $5$ . Dividing both sides by $2$ , we find that $-2 = a(1)^2 + b(1) + 2$ $7$ . Substituting $-2 = a(1)^2 + b(1) + 2$ $8$ back into the first equation, we get $-2 = a(1)^2 + b(1) + 2$ $9$ . Multiplying $2$ across to get rid of the fraction, we have $-2 = a + b + 2$ $1$ .