NOT NECESSARILY would a triangle be equilateral if one of its angles is 60 degrees. To be an equilateral triangle (a triangle in which all 3 sides have the same length), all 3 angles of the triangle would have to be 60°-angles; however, the triangle could be a 30°-60°-90° right triangle in which the side opposite the 30 degree angle is one-half as long as the hypotenuse, and the length of the side opposite the 60 degree angle is √3/2 as long as the hypotenuse. Another of possibly many examples would be a triangle with angles of 60°, 40°, and 80° which has opposite sides of lengths 2, 1.4845 (rounded to 4 decimal places), and 2.2743 (rounded to 4 decimal places), respectively, the last two of which were determined by using the Law of Sines: "In any triangle ABC, having sides of length a, b, and c, the following relationships are true: a/sin A = b/sin B = c/sin C."¹
X-15+9-4=x-10 there are ten fewer people after the second stop than the original amount
Answer:
your answers obviously 210
Hello from MrBillDoesMath!
Answer:
Top line: y = (2/3)x + 2
Bottom line: y = (2/3)x -1
Discussion:
The graph provided is hard to read but I did the best I could.
The top line appears to pass through the points (0,2) and (-3,0)
For this line
m = change y /change x = (0-2)/(-3-0) = -2/-3 = +2/3. So
y = mx + b => y = (2/3) x+ b. As the line passes through (0,2) set x = 0, y= 2 in y = (2/3)x + b =>
2 = (2/3) 0 + b => b = 2
Therefore y = (2/3)x + 2
The bottom line appears to pass through the points (0,-1) and (3,1)
For this line
m = change y /change x = (1-(-1)) /(3-0) = +2/-3. So
y = mx + b => y = (2/3) x+ b. As the line passes through (0,-1) set x = 0, y= -1 in y = (2/3)x + b =>
-1 = (2/3) 0 + b => b = -1
Therefore y = (2/3)x + -1
Thank you,
MrB
