Answer:
1, 2, -3, 51
Step-by-step explanation:
0 ÷ 2 = 0
0 + 1 = 1
2 ÷ 2 = 1
1 + 1 = 2
-8 ÷ 2 = -4
-4 + 1 = -3
100 ÷ 2 = 50
50 + 1 = 51
I’m not sure but this is what I would do
Runner A (unit rate / constant of proportionality):
54 feet divided by 4.5 seconds = 12 feet per second
Runner B (unit rate / constant of proportionality): 300 feet divided by 20 seconds (It looks like the end point of the line lands between 300 feet and 20 seconds) = 15 feet per second
Unit Rate:
Runner A = 12 feet per second
Runner B = 15 feet per second
Faster Runner:
Runner B is the faster runner as he can run a further distance than Runner A in one second.
Answer:
The time would be 4:59
Step-by-step explanation:
Ok so defenitions
isosceles means 2 angles or sides are equal legnth
scaleen means no sides or angles are eaqual measure
equilateral means all sides and angles are equal measure
(since all side equal measure, all angles are always 60)
acute means all angles are less than 90
obtuse means at least one angle is more than 90
right means that 1 angle is 90 degrees
ok so
isosceleese
for it to be acute, maybe make the angles 50,50,80
obtuse 40,40,100
right 90,45,45
scaleene
acute, 30,70,80
obtuse 100,30,50
right, 90,40,50
equilateral all angles are automatically equal to 60 so
acute: 60,60,60
obtuse: not possible
right: not possible
Answer:
The first step is to place the compass needle on one of the ends of the given segment AB, and draw an arc above and below the segment AB
Step-by-step explanation:
The steps to construct a perpendicular to segment AB are;
1) Place the compass needle on one of the ends of the given segment AB, and draw an arc above and below the segment AB
2) With the same compass width, place the compass needle on the other end of segment AB, and draw an arc above and below the segment AB intersecting the arcs constructed in step 1 above. Label the point of intersection as C and D
3) With a straight edge, draw a line joining the points of intersection C and D of the two arcs. The line CD. is the perpendicular bisector of the segment AB.
