If points f and g are symmetric with respect to the line y=x, then the line connecting f and g is perpendicular to y=x, and f and g are equidistant from y=x.
This problem could be solved graphically by graphing y=x and (8,-1). With a ruler, measure the perpendicular distance from y=x of (8,-1), and then plot point g that distance from y=x in the opposite direction. Read the coordinates of point g from the graph.
Alternatively, calculate the distance from y=x of (8,-1). As before, this distance is perpendicular to y=x and is measured along the line y= -x + b, where b is the vertical intercept of this line. What is b? y = -x + b must be satisfied by (8,-1): -1 = -8 + b, or b = 7. Then the line thru (8,-1) perpendicular to y=x is y = -x + 7. Where does this line intersect y = x?
y = x = y = -x + 7, or 2x = 7, or x = 3.5. Since y=x, the point of intersection of y=x and y= -x + 7 is (3.5, 3.5).
Use the distance formula to determine the distance between (3.5, 3.5) and (8, -1). This produces the answer to this question.
2/5 is in the lowest terms or simplest form
There are a few ways to go about doing this, but the easiest is portably to break it up into three shapes: two identical triangles, and a rectangle. The area of one triangle is:
1/2 (10x8) = 40, so the area of two triangles is 80.
The area of the square is 20 x 8, which is 160.
The total area is the sum of the two:
80 + 160 = 240, and this is your total area.
Answer:
A) b= 8 students
B) 1) 19/55 2) 28/55
Step-by-step explanation:
I'm sorry if I'm getting this whole concept wrong cause it felt way too easy to be right so I'm sorry if you get it wrong.
The reason the first question is 8 is because of the fact that you first have to add the students charted on the Venn-diagram and subtract that from 55, getting you to 8 students.
For the second one, the first part would be 19/55 becuase in the Venn-diagram it is shown that only 19 people passed both and so that be the numerator and the denominator would be the total of students, so it would be 55.
Same thing for part two except the fact that you have to add the students who only passed one subject, so combine 17 and 11, getting you to 28, thus getting the answer 28/55.
