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3 years ago

On a map, the distance between two

1 answer:

5 0

131.25 miles.

If 1 inch actually equals 25 inches, then you can just multiply 5.25 inches by 25, and because 1 in is equivalent to 25 miles, then your final answer will be in miles

You might be interested in

What is the range of Variable A for the cluster?

katovenus [111]

1. The range is the possible values for the y-coordinate in a relation or a set of data pairs.
The range is {nine, ten, eight, five, four, two}

2. If you plot the given points, then you'd see that that point
(three, nine) is far from the rest of the other points which makes it an outlier. Other methods can be used to determine whether or not the point is really and outlier. For now, a visual inspection is enough.

5 0

3 years ago

Mrs. Williamson had 18 students in her homeroom at the beginning of the school year and 24 students at the end of the year. What

jarptica [38.1K]

Answer:

25%

Step-by-step explanation:

An easy way to figure this out is just by dividing the initial amount of students (18) by the current amount of students (24), and then subtracting 1.

So, 18/24 = 0.75 (aka 75%), and 1 - 0.75 = 0.25 (aka 25%)

7 0

2 years ago

A certain square is to be drawn on a coordinate plane. One of the vertices must be on the origin, and the square is to have an a

Scrat [10]

Answer:

The answer is (C) 8

Step-by-step explanation:

First, let's calculate the length of the side of the square.

$A_{square}=a^2$ , where $a$ is the length of the side. Now, let's try to build the square. First we need to find a point which distance from (0, 0) is 10. For this, we can use the distance formula in the plane:

$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ which for $x_1=0$ and $y_1 = 0$ transforms as $d=\sqrt{(x_2)^2 + (y_2)^2}$ . The first point we are looking for is connected to the origin and therefore, its components will form a right triangle in which, the Pythagoras theorem holds, see the first attached figure. Then, $x_2$ , $y_2$ and 10 are a Pythagorean triple. From this, $x_2= 6$ or $x_2=8$ while $y_2= 6$ or $y_2=8$ . This leads us with the set of coordinates:

$(\pm 6, \pm 8)$ and $(\pm 8, \pm 6)$ . (A)

The next step is to find the coordinates of points that lie on lines which are perpendicular to the lines that joins the origin of the coordinate system with the set of points given in (A):

Let's do this for the point (6, 8).

The equation of the line that join the point (6, 8) with the origin (0, 0) has the equation $y = mx +n$ , however, we only need to find its slope in order to find a perpendicular line to it. Thus,

$m = \frac{y_2-y_1}{x_2-x_1} \\m = \frac{8-0}{6-0} \\m = 8/6$

Then, a perpendicular line has an slope $m_{\bot} = -\frac{1}{m} = -\frac{6}{8}$ (perpendicularity condition of two lines). With the equation of the slope of the perpendicular line and the given point (6, 8), together with the equation of the distance we can form a system of equations to find the coordinates of two points that lie on this perpendicular line.

$m_{\bot}=\frac{6}{8} = \frac{8-y}{6-x}\\ 6(6-x)+8(8-y)=0$ (1)

$d^2 = \sqrt{(y_o-y)^2+(x_o-x)^2} \\(10)^2=\sqrt{(8-y)^2+(6-x)^2}\\100 = \sqrt{(8-y)^2+(6-x)^2}$ (2)

This system has solutions in the coordinates (-2, 14) and (14, 2). Until here, we have three vertices of the square. Let's now find the fourth one in the same way we found the third one using the point (14,2). A line perpendicular to the line that joins the point (6, 8) and (14, 2) has an slope $m = 8/6$ based on the perpendicularity condition. Thus, we can form the system:

$\frac{8}{6} =\frac{2-y}{14-x} \\8(14-x) - 6(2-y) = 0$ (1)

$100 = \sqrt{(14-x)^2+(2-y)^2}$ (2)

with solution the coordinates (8, -6) and (20, 10). If you draw a line joining the coordinates (0, 0), (6, 8), (14, 2) and (8, -6) you will get one of the squares that fulfill the conditions of the problem. By repeating this process with the coordinates in (A), the following squares are found:

(0, 0), (6, 8), (14, 2), (8, -6)

(0, 0), (8, 6), (14, -2), (6, -8)
(0, 0), (-6, 8), (-14, 2), (-8, -6)
(0, 0), (-8, 6), (-14, -2), (-6, -8)

Now, notice that the equation of distance between the two points separated a distance of 10 has the trivial solution $(\pm10, 0)$ and $(0, \pm10)$ . By combining this points we get the following squares: