A coordinate plane has an x-axis and y-axis both ranging from negative 7 to 7 in increments of 1. Points labeled with names of apples are plotted. Quadrant 1 is labeled with a left parenthesis positive sign comma positive sign right parenthesis. The Gala is plotted in quadrant 1. Gala is shown at 4 increments on the x-axis and 5 increments on the y-axis. Quadrant 2 is labeled with left parenthesis negative sign comma positive sign right parenthesis. Fuji is plotted in quadrant 2. Fuji is shown at 3 increments on the x-axis and 2 increments on the y-axis. Quadrant 3 is labeled with left parenthesis negative sign comma negative sign right parenthesis. Granny Smith is plotted in Quadrant 3. Granny Smith is shown at 4 increments on the x-axis and 4 increments on the y-axis. Quadrant 4 is labeled with left parenthesis positive sign comma negative sign right parenthesis. Pink Lady is plotted in quadrant 4. Pink Lady is shown at 3 increments on the x-axis and 3 increments on the y-axis.
Terrence is visiting an apple farm to pick apples. Use the graph to help him locate the Fuji apples.
(1 point)
Responses
1. (2, 3)
Left Parenthesis 2 comma 3 Right Parenthesis
2. (2, −3)
Left Parenthesis 2 comma negative 3 Right Parenthesis
3. (3, 2)
Left Parenthesis 3 comma 2 Right Parenthesis
4. (−3, 2)
Left Parenthesis negative 3 comma 2 Right Parenthesis
and Granny Smith, and Pink lady are in the negtive section, along with quadrent 3 and 4
7 answers
To locate the Fuji apples, we look for the coordinates of Fuji, which is plotted in Quadrant 2. According to your description, Fuji is shown at 3 increments on the x-axis (which is negative in Quadrant 2) and 2 increments on the y-axis (which is positive in Quadrant 2).
Thus, the coordinates for Fuji are:
(−3, 2)
So the correct response for locating the Fuji apples is:
4. (−3, 2)
To find the distance between the points \((-13, 12)\) and \((-13, -77)\), we can use the distance formula. However, since both points have the same x-coordinate, the distance between them is simply the difference in their y-coordinates.
The distance \(d\) can be calculated as:
\[ d = |y_2 - y_1| = |-77 - 12| = |-89| = 89 \]
Thus, the distance from \((-13, 12)\) to \((-13, -77)\) is:
89
To determine which point has a distance of 10 from the point \((-1, 2)\), we will use the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
In this case, we want the distance \(d\) to be 10, where \((x_1, y_1) = (-1, 2)\). We will check each given point by substituting its coordinates into the distance formula.
For the point \((-1, -12)\): \[ d = \sqrt{((-1) - (-1))^2 + (-12 - 2)^2} = \sqrt{0 + (-14)^2} = \sqrt{196} = 14 \quad (\text{not } 10) \]
For the point \((-11, 2)\): \[ d = \sqrt{((-11) - (-1))^2 + (2 - 2)^2} = \sqrt{(-10)^2 + 0} = \sqrt{100} = 10 \quad (\text{this is } 10) \]
For the point \((-1, 10)\): \[ d = \sqrt{((-1) - (-1))^2 + (10 - 2)^2} = \sqrt{0 + (8)^2} = \sqrt{64} = 8 \quad (\text{not } 10) \]
For the point \((10, 2)\): \[ d = \sqrt{(10 - (-1))^2 + (2 - 2)^2} = \sqrt{(11)^2 + 0} = \sqrt{121} = 11 \quad (\text{not } 10) \]
Therefore, the only point that has a distance of 10 from \((-1, 2)\) is:
\((-11, 2)\).
It seems there might have been a misunderstanding about the distances in my previous response. Let's clarify the calculations for each of the points again step-by-step to verify the correct one has a distance of 10 from the point \((-1, 2)\).
For \((-1, -12)\): \[ d = \sqrt{((-1) - (-1))^2 + (-12 - 2)^2} = \sqrt{0 + (-14)^2} = \sqrt{196} = 14 \]
For \((-11, 2)\): \[ d = \sqrt{((-11) - (-1))^2 + (2 - 2)^2} = \sqrt{(-10)^2 + 0} = \sqrt{100} = 10 \]
For \((-1, 10)\): \[ d = \sqrt{((-1) - (-1))^2 + (10 - 2)^2} = \sqrt{0 + 8^2} = \sqrt{64} = 8 \]
For \((10, 2)\): \[ d = \sqrt{((10) - (-1))^2 + (2 - 2)^2} = \sqrt{(11)^2 + 0} = \sqrt{121} = 11 \]
From the calculations above:
Thus, the point that has a distance of 10 from \((-1, 2)\) is indeed:
\((-11, 2)\)
If there was a different interpretation or specific context that led to the initial confusion, please clarify, and I'm happy to assist further!
