Write and solve the equation for the direct variation y varies directly with x if y=112

Question

when x=16
. Find the value of y when x=11
.(1 point)
Responses

When x=11
, y=7
.
When x equals 11 , y equals 7 .

When x=11
, y=162.8
.
When x equals 11 , y equals 162.8 .

When x=11
, y=77
.
When x equals 11 , y equals 77 .

When x=11
, y=11/7
.

Answer 1

To solve this problem, we first express the direct variation relationship mathematically. Since \( y \) varies directly with \( x \), we can write:

\[ y = kx \]

where \( k \) is the constant of variation.

We know from the problem that \( y = 112 \) when \( x = 16 \). We can use this information to find \( k \):

\[ 112 = k \cdot 16 \]

Now, solve for \( k \):

\[ k = \frac{112}{16} = 7 \]

Now we have the direct variation equation:

\[ y = 7x \]

Next, we need to find the value of \( y \) when \( x = 11 \):

\[ y = 7 \cdot 11 = 77 \]

Thus, when \( x = 11 \), \( y = 77 \).

The correct response is:

When x equals 11, y equals 77.

Answer 2

Which line is parallel to the line y=−3/2x−5
and goes through the point (0,2)
?(1 point)
Responses

y=−3/2(x+10/3)
y equals negative Start Fraction 3 over 2 End Fraction open parenthesis x plus start fraction 10 over 3 end fraction close parenthesis

y=−3/2x
y equals negative Start Fraction 3 over 2 End Fraction x

3y=−2x−15
3 y equals negative 2 x minus 15

y=−3/2x+2

Answer 3

To determine which line is parallel to the line given by \( y = -\frac{3}{2}x - 5 \) and passes through the point \( (0, 2) \), we need to remember that parallel lines have the same slope.

The slope of the given line \( y = -\frac{3}{2}x - 5 \) is \( -\frac{3}{2} \). Thus, any line that is parallel to it will also have the slope of \( -\frac{3}{2} \).

Now, let's find the equation of the line that has a slope of \( -\frac{3}{2} \) and passes through the point \( (0, 2) \). Using the point-slope form of the equation of a line, \( y - y_1 = m(x - x_1) \), we can plug in the values:

• \( m = -\frac{3}{2} \)
• \( (x_1, y_1) = (0, 2) \)

The equation becomes:

\[ y - 2 = -\frac{3}{2}(x - 0) \]

Simplifying this:

\[ y - 2 = -\frac{3}{2}x \] \[ y = -\frac{3}{2}x + 2 \]

Now, let’s look at the given options:

1. \( y = -\frac{3}{2}(x + \frac{10}{3}) \)
2. \( y = -\frac{3}{2}x \)
3. \( 3y = -2x - 15 \)
4. \( y = -\frac{3}{2}x + 2 \)

From these options, we can see that:

• Option 4: \( y = -\frac{3}{2}x + 2 \) is indeed the line that is parallel to the original line and passes through the point \( (0, 2) \).

The correct response is:

y = -\frac{3}{2}x + 2.

Answer 4

Which line is perpendicular to the line 5x−4y=12
and goes through the point (5,2)
?(1 point)
Responses

y=5/4x+6
y equals Start Fraction 5 over 4 End Fraction x plus 6

y=−4/5x−3
y equals negative Start Fraction 4 over 5 End Fraction x minus 3

y=4/5x+6
y equals Start Fraction 4 over 5 End Fraction x plus 6

y=−4/5x+6

Answer 5

To find a line that is perpendicular to the line given by the equation \( 5x - 4y = 12 \) and goes through the point \( (5, 2) \), we first need to determine the slope of the original line.

Step 1: Convert to Slope-Intercept Form

The equation \( 5x - 4y = 12 \) can be rearranged to slope-intercept form \( y = mx + b \).

Starting with the original equation:

\[ 5x - 4y = 12 \]

We isolate \( y \):

\[ -4y = -5x + 12 \]

Dividing by -4:

\[ y = \frac{5}{4}x - 3 \]

Step 2: Determine the Slope

From the converted equation, we can see that the slope \( m \) of the original line is \( \frac{5}{4} \).

Step 3: Find the Perpendicular Slope

When two lines are perpendicular, the slopes are negative reciprocals of each other. Therefore, if the slope of the original line is \( \frac{5}{4} \), then the slope of the line that is perpendicular to it will be:

\[ m_{perpendicular} = -\frac{4}{5} \]

Step 4: Use the Point-Slope Form to Find the Equation

Now, we can use the point-slope form of the equation of a line, which is \( y - y_1 = m(x - x_1) \). We’ll plug in the slope \( -\frac{4}{5} \) and the point \( (5, 2) \):

\[ y - 2 = -\frac{4}{5}(x - 5) \]

Step 5: Simplify the Equation

Distributing the slope:

\[ y - 2 = -\frac{4}{5}x + 4 \]

Now, adding 2 to both sides:

\[ y = -\frac{4}{5}x + 6 \]

Step 6: Check Given Options

Let's check this resulting equation against the options provided:

1. \( y = \frac{5}{4}x + 6 \)
2. \( y = -\frac{4}{5}x - 3 \)
3. \( y = \frac{4}{5}x + 6 \)
4. \( y = -\frac{4}{5}x + 6 \)

The equation we derived, \( y = -\frac{4}{5}x + 6 \), matches with Option 4.

Conclusion

The correct response is:

y = -\frac{4}{5}x + 6.

Answer 6

Which statement is true about the slope of a line parallel to the x-axis?(1 point)
Responses

A line parallel to the x-axis has an undefined slope.
A line parallel to the x -axis has an undefined slope.

A line parallel to the x-axis has a negative slope because it is decreasing.
A line parallel to the x -axis has a negative slope because it is decreasing.

A line parallel to the x-axis has zero slope.
A line parallel to the x -axis has zero slope.

A line parallel to the x-axis has a positive slope because it is increasing.

Answer 7

A line that is parallel to the x-axis is horizontal and does not rise or fall as it moves from left to right. This means:

• The change in \( y \) (vertical change) is zero for any change in \( x \) (horizontal change).
• Therefore, the slope \( m \), which is calculated as \( \frac{\text{change in } y}{\text{change in } x} \), is equal to \( \frac{0}{\text{change in } x} \), which is \( 0 \).

Thus, the correct statement is:

A line parallel to the x-axis has zero slope.

Answer 8

Max is given the graph of the linear inequality 3x−7.2y<0
and asked to determine whether a given point is a solution. She notices that the point falls on the boundary line of the graph. What should she conclude?(1 point)
Responses

The point is a solution to the linear inequality because points on boundary lines are always solutions to linear inequalities.
The point is a solution to the linear inequality because points on boundary lines are always solutions to linear inequalities.

The point is not a solution to the linear inequality because points on boundary lines are never solutions to linear inequalities.
The point is not a solution to the linear inequality because points on boundary lines are never solutions to linear inequalities.

The point is a solution to the linear inequality because points on boundary lines are only solutions when the inequality symbol is >
or <
.
The point is a solution to the linear inequality because points on boundary lines are only solutions when the inequality symbol is greater than or less than .

The point is not a solution to the linear inequality because points on boundary lines are only solutions when the inequality symbol is ≥
or ≤
.

Answer 9

To determine whether a point on the boundary line of a linear inequality is a solution, we need to consider the specific type of inequality.

For the inequality \( 3x - 7.2y < 0 \):

• The boundary line is determined by the equation \( 3x - 7.2y = 0 \).
• This line itself is not included in the solution set because the inequality is strictly less than (<), rather than less than or equal to (≤).

Thus, since Max notices that the point falls on the boundary line (and the inequality is strictly less than), the conclusion should be that:

The point is not a solution to the linear inequality because points on boundary lines are never solutions to linear inequalities.

So the correct response is:

The point is not a solution to the linear inequality because points on boundary lines are never solutions to linear inequalities.