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Which equation represents a line that passes through (4, left-parenthesis 4, StartFraction one-third EndFraction right-parenthesis.) and has a slope of StartFraction 3 Over 4 EndFraction.?

y – y minus StartFraction one-third EndFraction equals StartFraction 3 Over 4 EndFraction left-parenthesis x minus 4 right-parenthesis. = y minus StartFraction 3 Over 4 EndFraction equals StartFraction one-third EndFraction left-parenthesis x minus 4 right-parenthesis.(x – 4)
y – y minus StartFraction one-third EndFraction equals 4 left-parenthesis x minus StartFraction 3 Over 4 EndFraction right-parenthesis. = y minus StartFraction one-third EndFraction equals StartFraction 3 Over 4 EndFraction left-parenthesis x minus 4 right-parenthesis.(x – 4)
y – y minus StartFraction one-third EndFraction equals 4 left-parenthesis x minus StartFraction 3 Over 4 EndFraction right-parenthesis. = 4(x – )
y – 4 = y minus 4 equals StartFraction 3 Over 4 EndFraction left-parenthesis x minus StartFraction one-third EndFraction right-parenthesis.(x – )

2. Talia wants to write the equation of the graphed line in point-slope form. These are the steps she plans to use:Step 1: Choose a point on the line, such as (2, 5).Step 2: Choose another point on the line, such as (1, 3).Step 3: Count units to determine the slope ratio. The line runs 1 unit to the right and rises 2 units up, so the slope is StartFraction one-half EndFraction.Step 4: Substitute those values into the point-slope form.y – y1 = m(x – x1)y – 3 = y minus y 1 equals m left-parenthesis x minus x 1 right-parenthesis. (x – 1)Which of Talia’s steps is incorrect?Step 1 is incorrect because it uses a point that is not on the line.Step 2 is incorrect because it uses a point that is not on the line.Step 3 is incorrect because it shows an incorrect ratio for the slope.Step 4 is incorrect because it shows an incorrect substitution of (1, 3) into the point-slope form.

3. Which equation shows the point-slope form of the line that passes through (3, 2) and has a slope of y plus StartFraction one-half EndFraction equals 3 left-parenthesis x minus 2 right-parenthesis.?

y + 2 =y plus 2 equals StartFraction one-third EndFraction left-parenthesis x plus 3 right-parenthesis.(x + 3)
y – 2 = y minus 2 equals StartFraction one-third EndFraction left-parenthesis x minus 3 right-parenthesis.(x – 3)
y + 3 = y plus 3 equals StartFraction one-third EndFraction left-parenthesis x plus 2 right-parenthesis.(x + 2)
y – 3 = y plus StartFraction one-half EndFraction equals 2 left-parenthesis x minus 3 right-parenthesis.(x – 2)

4. Chin was shown the graph of a line that contained point (1, 7). He wrote f(x) = 4x + 3 to correctly represent the line. Which of these equations could represent the same line?

y – 7 = 3(x – 1)
y – 1 = 3(x – 7)
y – 7 = 4(x – 1)
y – 1 = 4(x – 7)

5. What is the slope of the line whose equation is y−4= StartFraction 5 Over 2 EndFraction.(x−2)?A coordinate plane with a line passing through the points (0, negative 1) and (2, 4).–negative StartFraction 5 Over 2 EndFraction.–negative StartFraction 2 Over 5 EndFraction.StartFraction 2 Over 5 EndFraction.StartFraction 5 Over 2 EndFraction.

6. Which equations and/or functions represent the graphed line? Select three options.A coordinate plane with a line passing through the points, (negative 4, 0), (negative 2, 1), (0, 2), and (2, 3).y – 3 = y minus 3 equals StartFraction one-half EndFraction left-parenthesis x minus 2 right-parenthesis.(x – 2)

7. Which equation represents a line that passes through (5, 1) and has a slope of StartFraction one-half EndFraction?

y – 5 = y minus 5 equals StartFraction one-half EndFraction left-parenthesis x minus 1 right-parenthesis.(x –1)
y – y minus StartFraction one-half EndFraction equals 5 left-parenthesis x minus 1 right-parenthesis. = 5(x –1)
y – 1 = y minus 1 equals StartFraction one-half EndFraction left-parenthesis x minus 5 right-parenthesis.(x –5)
y – 1 = 5y minus 1 equals 5 left-parenthesis x minus StartFraction one-half EndFraction right-parenthesis.

8. The equation of a linear function in point-slope form is y – y1 = m(x – x1). Harold correctly wrote the equation y = 3(x – 7) using a point and the slope. Which point did Harold use?

(7, 3)
(0, 7)
(7, 0)
(3, 7)

9. Anya graphed the line (y−2)=3(x−1) on the coordinate grid.A coordinate plane with a line passing through the points, (negative 2, negative 7), (0, negative 1), and (1, 2).What is the slope of Anya’s line?−3−113

10. Which equation represents a line that passes through (–2, 4) and has a slope of StartFraction 2 Over 5 EndFraction.?

9 answers

Let's go through each of your questions one by one, providing the correct answers and explanations.

Question 1:

Which equation represents a line that passes through \((4, \frac{1}{3})\) and has a slope of \(\frac{3}{4}\)?

We use the point-slope form of the equation of a line, which is: \[y - y_1 = m(x - x_1)\]

Here, \((x_1, y_1) = (4, \frac{1}{3})\) and \(m = \frac{3}{4}\). Thus, the equation is: \[y - \frac{1}{3} = \frac{3}{4}(x - 4)\]

Question 2:

Talia's steps: Which of her steps is incorrect?

  • Step 3 is incorrect because the slope she calculated is \( \frac{2}{1} = 2\) (rises 2 units and runs 1 unit), hence the slope should be 2, not \(\frac{1}{2}\).

Question 3:

Which equation shows the point-slope form of the line that passes through (3, 2) and has a slope of \(\frac{1}{3}\)?

Using \((x_1, y_1) = (3, 2)\) and \(m = \frac{1}{3}\): \[y - 2 = \frac{1}{3}(x - 3)\]

None of the given options matches this point-slope form. The correct answer should be: \[y - 2 = \frac{1}{3}(x - 3)\]

Question 4:

Which of these equations could represent the same line?

Since the slope from \(f(x) = 4x + 3\) is 4 and it passes through (1, 7), we need an equation with the same point and slope.

The correct equations are:

  • \(y - 7 = 4(x - 1)\).

Question 5:

What is the slope of the line whose equation is \(y−4=\frac{5}{2}(x−2)\)?

The equation is already in point-slope form. The slope is:

  • \(\frac{5}{2}\).

Question 6:

Which equations and/or functions represent the graphed line?

To determine the representation, you would typically evaluate the slope and a point from the graph. However, without specific options, we can't determine which three options represent the line. The valid options would typically reflect that slope.

Question 7:

Which equation represents a line that passes through \((5, 1)\) and has a slope of \(\frac{1}{2}\)?

Using \((x_1, y_1) = (5, 1)\) and \(m = \frac{1}{2}\): \[y - 1 = \frac{1}{2}(x - 5)\]

Question 8:

Which point did Harold use?

Using the point-slope equation \(y - y_1 = m(x - x_1)\) with \(y = 3(x - 7)\), it can be understood he used:

  • (7, 21), given he wrote the equation derived based on that point.

Question 9:

What is the slope of Anya’s line?

The equation is in point-slope form: \(y−2=3(x−1)\), hence the slope \(m\) is:

  • 3.

Question 10:

Which equation represents a line that passes through \((-2, 4)\) and has a slope of \(\frac{2}{5}\)?

Using \((x_1, y_1) = (-2, 4)\) and \(m = \frac{2}{5}\): \[y - 4 = \frac{2}{5}(x + 2)\]

This gives us the point-slope form for the specified point and slope.

Let me know if you have any further questions!

for 6 and 8

6. A coordinate plane with a line passing through the points, (negative 4, 0), (negative 2, 1), (0, 2), and (2, 3).y – 3 = y minus 3 equals StartFraction one-half EndFraction left-parenthesis x minus 2 right-parenthesis.(x – 2)
Which equations and/or functions represent the graphed line? Select three options.

y – 3 = y minus 3 equals StartFraction one-half EndFraction left-parenthesis x minus 2 right-parenthesis.(x – 2)

and 8. The equation of a linear function in point-slope form is y – y1 = m(x – x1). Harold correctly wrote the equation y = 3(x – 7) using a point and the slope. Which point did Harold use?

(7, 3)
(0, 7)
(7, 0)
(3, 7)

only those options are available

Thank you for providing the additional context. Let's address both questions with the provided options.

Question 6:

A coordinate plane with a line passing through the points (–4, 0), (–2, 1), (0, 2), and (2, 3). Which equations and/or functions represent the graphed line?

You mentioned an option:

  • \(y - 3 = \frac{1}{2}(x - 2)\)

First, let's determine the slope of the line that passes through the given points. The points you've provided seem to suggest a line that increases steadily.

  1. To find the slope between any two points:
    • Between \((-4, 0)\) and \((-2, 1)\): \[ m = \frac{1 - 0}{-2 + 4} = \frac{1}{2} \]
    • Similarly, calculating between \((-2, 1)\) and \((0, 2)\): \[ m = \frac{2 - 1}{0 + 2} = \frac{1}{2} \]
    • And for \((0, 2)\) and \((2, 3)\): \[ m = \frac{3 - 2}{2 - 0} = \frac{1}{2} \]

Thus, the slope \(m\) is indeed \(\frac{1}{2}\).

Now the equation in point-slope form can be revised for one of the points, like \((2, 3)\): \[ y - 3 = \frac{1}{2}(x - 2) \]

From the options, you should look for any variation of the line with a slope of \(\frac{1}{2}\) that uses one of the points through which the line passes. The key will be considering the related points like \((-4, 0)\) or any other options that yield similar equations.

Since you only had provided one option, please check whether it was indeed the only option available that matches the slope.

Question 8:

The equation of a linear function in point-slope form is \(y - y_1 = m(x - x_1)\). Harold correctly wrote the equation \(y = 3(x - 7)\) using a point and the slope. Which point did Harold use?

The equation provided tells us he used the point-slope form, where \(m = 3\) and the point is \((x_1, y_1) = (7, y_1)\).

To determine \(y_1\), we can evaluate it:

  • The point-slope form states: \[ y - y_1 = 3(x - 7) \]

If he wrote the equation as \(y = 3(x - 7)\), he essentially set \(y_1\) to \(0\) when rearranged to the y-intercept form (since at \(x = 7\), \(y\) also equals 0 when substituting back).

Thus, the point used by Harold is:

  • (7, 0).

If you only have the options available:

  • (7, 3)
  • (0, 7)
  • (7, 0) ← correct point
  • (3, 7)

The answer is clearly (7, 0).

Let me know if you need additional clarification or further assistance!

im still confused about 6. Which equations and/or functions represent the graphed line? Select three options.A coordinate plane with a line passing through the points, (negative 4, 0), (negative 2, 1), (0, 2), and (2, 3).

these are the options
f(x)=1/5x-4
f(x)=1/2x+2
f(x)=1/2x+1
y-3=1/2(x-2)
y-1=1/2(x+2)

Sure! Let's determine which of the provided equations correspond to the line that passes through the points \((-4, 0)\), \((-2, 1)\), \((0, 2)\), and \((2, 3)\).

Step 1: Calculate the Slope

We already found earlier that the slope of this line is \( \frac{1}{2} \). Let’s check it again using two of the given points:

  1. Between \((-4, 0)\) and \((-2, 1)\): \[ m = \frac{1 - 0}{-2 + 4} = \frac{1}{2} \]

  2. Between \((-2, 1)\) and \((0, 2)\): \[ m = \frac{2 - 1}{0 + 2} = \frac{1}{2} \]

  3. Between \((0, 2)\) and \((2, 3)\): \[ m = \frac{3 - 2}{2 - 0} = \frac{1}{2} \]

Step 2: Find the Equation of the Line

The line has a slope of \( \frac{1}{2} \). We can write the equation of the line using point-slope form.

We can use any of the points. Let’s use the point \((0, 2)\):

Using the point-slope form: \[ y - y_1 = m(x - x_1) \] Substituting \(m = \frac{1}{2}\) and \((x_1, y_1) = (0, 2)\): \[ y - 2 = \frac{1}{2}(x - 0) \] This simplifies to: \[ y - 2 = \frac{1}{2}x \quad \text{or} \quad y = \frac{1}{2}x + 2 \]

Step 3: Compare to Given Options

Now, let’s look at the options you provided:

  1. \(f(x) = \frac{1}{5}x - 4\) (not correct, wrong slope)
  2. \(f(x) = \frac{1}{2}x + 2\) (correct!)
  3. \(f(x) = \frac{1}{2}x + 1\) (not correct, wrong y-intercept)
  4. \(y - 3 = \frac{1}{2}(x - 2)\) (correct! Because it can be rearranged and will yield y-intercept of 2)
  5. \(y - 1 = \frac{1}{2}(x + 2)\) (not correct, as it gives \(y\) intercept of 1)

Conclusion

The equations that represent the line passing through the given points are:

  • \(f(x) = \frac{1}{2}x + 2\)
  • \(y - 3 = \frac{1}{2}(x - 2)\)

Now, we need three expressions. Since we have identified only two from your provided options, we can look for an additional function or confirm that other forms (like manipulations of the second option).

Thus, the two confirmed equations are correct, but only two are clearly provided here and none can replicate a third that fits with accuracy based on the nature of pure linear transformation.

To summarize, your chosen equations are:

  • \(f(x) = \frac{1}{2}x + 2\)
  • \(y - 3 = \frac{1}{2}(x - 2)\)

No valid third option appears present from the stated answers provided without further equivalent adjustments.

If you have any further queries, I’d be happy to assist!

well the question asks for 3 so pick the best option

I apologize for the oversight. Since we need to select three equations that represent the line passing through the given points, let's analyze all the provided options more closely.

Given Options:

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

Analyzing Each Option:

  1. \(f(x) = \frac{1}{5}x - 4\): This has a slope of \(\frac{1}{5}\), which is incorrect. (Not valid)

  2. \(f(x) = \frac{1}{2}x + 2\): This is correct as it matches our derived equation. (Valid)

  3. \(f(x) = \frac{1}{2}x + 1\): This has a y-intercept of 1, which does not match the line we found. (Not valid)

  4. \(y - 3 = \frac{1}{2}(x - 2)\): Rearranging gives \(y - 3 = \frac{1}{2}x - 1\) or \(y = \frac{1}{2}x + 2\). This matches the correct line as well. (Valid)

  5. \(y - 1 = \frac{1}{2}(x + 2)\): Rearranging gives \(y - 1 = \frac{1}{2}x + 1\) or \(y = \frac{1}{2}x + 2\). This does not match either, as it gives a y-intercept of 2 with the correct slope but incorrectly representing the offset from the original point. (Not valid)

Selected Valid Equations:

From the options:

  1. \(f(x) = \frac{1}{2}x + 2\) (Valid)
  2. \(y - 3 = \frac{1}{2}(x - 2)\) (Valid)

Selecting a Third Option:

Since only two valid representations can be found in the options, let's provide a reasonable alternative from the available choices:

Since \(y - 1 = \frac{1}{2}(x + 2)\) might still represent a line with a slope of \(\frac{1}{2}\), we’ll include it for the sake of finding a function with a similar slope even if it does not appear on the same line.

Final Selections:

  1. \(f(x) = \frac{1}{2}x + 2\)
  2. \(y - 3 = \frac{1}{2}(x - 2)\)
  3. \(y - 1 = \frac{1}{2}(x + 2)\) (Although it does not coincide with the given points, it's valid based on slope)

In conclusion, choose:

  • 1. \(f(x) = \frac{1}{2}x + 2\)
  • 2. \(y - 3 = \frac{1}{2}(x - 2)\)
  • 3. \(y - 1 = \frac{1}{2}(x + 2)\) (as a slope alternative)

Please let me know if there’s anything else you need!

YOU GOT ALMOST ALL CORRECT!! WOOO

1. y – 1/3 = 3/4(x-4)

Great to hear that we got almost everything correct! Thank you for confirming the point-slope form for question 1:

  1. The equation \(y - \frac{1}{3} = \frac{3}{4}(x - 4)\) indeed represents a line that passes through the point \((4, \frac{1}{3})\) with a slope of \(\frac{3}{4}\).

If you have any more questions or need further clarification on any other problems, feel free to ask! I'm here to help!