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Serhud [2]
3 years ago
7

By which angle must turn about point A in the clockwise direction so that it coincides with ?

Mathematics
1 answer:
schepotkina [342]3 years ago
5 0

When a line is rotated, it is rotated about a fixed point to its new position. Line AB must turn 180 degrees about point A to coincide with line AE.

<em>The complete question is that we determine the angle which AB will turn to, about point A to align with AE (see attachment for the figure)</em>

Line AB and AE form a straight line. This means that the angle between these two lines on either side of point A is 180^o

So for line AB to coincides with line AE, the line AB must turn 180 degrees about point A.

Hence, option (B) is correct.

Read more about rotation of points at:

brainly.com/question/22363471

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What’s 15 multiplied by 2/5 equals 6 in a division equation
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Answer:

6 / 15 = ?

Step-by-step explanation:

To put this in a division equation, you first need the total whivh is 6. So, to put this in a division equation it would look like 6 / 15 = ?

7 0
3 years ago
Use slopes to determine if the lines y= -x/3- 9 and 6x – 2y = -5 are perpendicular.
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Answer:

They are perpendicular.

Step-by-step explanation:

When we convert the equation 6x-2y=-5 into y=mx+b form, it changes into y=3x-5/2. Since we know that perpendicular slopes are opposite reciprocal, the opposite reciprocal of -x/3 is 3. So, we know that these 2 slopes are perpendicular.

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Who knows this answer!!
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Find the medion to 3,17,17,11,8,13,5,18
larisa [96]
The median of these numbers will be 12
 
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Read 2 more answers
Let F = (2, 3). Find coordinates for three points that are equidistant from F and the y-axis. Write an equation that says P = (x
True [87]

Answer:

The equation that says P is equidistant from F and the y-axis is P(x,y) =\left(1,\frac{3+y'}{2} \right).

(1, 0), (1, 3/2) and (1,6) are three points that are equdistant from F and the y-axis.

Step-by-step explanation:

Let F(x,y) = (2,3) and R(x,y) =(0, y'), where P(x,y) is a point that is equidistant from F and the y-axis. The following vectorial expression must be satisfied to get the location of that point:

F(x,y)-P(x,y) = P(x,y)-R(x,y)

2\cdot P(x,y) = F(x,y)+R(x,y)

P(x,y) = \frac{1}{2}\cdot F(x,y)+\frac{1}{2} \cdot R(x,y) (1)

If we know that F(x,y) = (2,3) and R(x,y) = (0,y'), then the resulting vectorial equation is:

P(x,y) = \left(1,\frac{3}{2} \right)+\left(0, \frac{y'}{2}\right)

P(x,y) =\left(1,\frac{3+y'}{2} \right)

The equation that says P is equidistant from F and the y-axis is P(x,y) =\left(1,\frac{3+y'}{2} \right).

If we know that y_{1}' = -3, y_{2}' = 0 and y_{3}' = 3, then the coordinates for three points that are equidistant from F and the y-axis:

P_{1}(x,y) = \left(1,\frac{3+y_{1}'}{2} \right)

P_{1}(x,y) = (1,0)

P_{2}(x,y) = \left(1,\frac{3+y_{2}'}{2} \right)

P_{2}(x,y) = \left(1,\frac{3}{2} \right)

P_{3}(x,y) = \left(1,\frac{3+y_{3}'}{2} \right)

P_{3}(x,y) = \left(1,6 \right)

(1, 0), (1, 3/2) and (1,6) are three points that are equdistant from F and the y-axis.

7 0
3 years ago
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