The given line segment has a midpoint at (3, 1). What is the equation, in slope-intercept form, of the perpendicular bisector of
the given line segment? y = x y = x – 2 y = 3x y = 3x − 8
2 answers:
Answer:
Step-by-step explanation:
I think the attached photo supports for your question
Here is my anser:
We need to find the slope of the of and from the graph, we see that if x increases from 2 to 4, y decreases from 4 to -2. Thus, the slope of the blue line is : = -3
But the slope of the perpend. bisector of the blue line is the negative reciprocal of -3, or m= 1/3.
Let's find the slope-intercept form of this bisector. We need to determine b in y=mx+b. Referring to the midpoint of the blue line, x= 3; y= 1; and m=1/3. Then
y=mx+b becomes 1=(1/3)(3) + b. Solving for b: 1=1+b. Then b=0.
Thus, the equation of the perpendicular bisector of the blue line through (3,1) is y=mx + b, or y=(1/3)x + 0, or y=x/3.
You might be interested in
Answer:
Area of the shape = 240 unit²
Step-by-step explanation:
Shape in the figure attached there are two parts. One is a triangle and other a rectangle.
By adding the area of these two shapes we can get the area of complete shape.
1). Area of the triangle =
Base of the triangle = 20 units
height of the triangle = 4 units
Therefore area of triangle = square units
2). Area of rectangle = Length × width
= 20 × 10 = 200 unit²
Total area of the shape = area of triangle + area of rectangle
= 40 + 200 = 240 unit²
So the answer is area of the shape = 240 unit²
Answer:
The correct option is 1.
Step-by-step explanation:
From the given histogram it is clear that
In age group 5-10, the number of students is 7.
In age group 11-13, the number of students is 5.
In age group 14-18, the number of students is 4.
It means age of 7 students lie in the interval 5-10, age of 5 students lie in the interval 11-13 and age of 4 students lie in the interval 14-18.
Total number of elements in the data set is
In option 3 and 4 number of elements is data set is less than 16. So, option 3 and 4 are incorrect.
In option 2 all the data set lie in the interval 5-10. So, option 2 is incorrect.
In option 1,
Class intervals elements frequency
5-10 5,6,7,8,9,10,10 7
11-13 11,11,12,12,13 5
14-18 14,14,15,16 4
From the above table we can conclude that data set 1 represented in the given histogram.
Therefore the correct option is 1.