8(7) = 56
7(6) = 42
6(5) = 30
5(4) = 20
4(3) = 12
3(2) = 6
∴ 3 = 6
We have to assume that the speed before being stuck was sufficient to get to the destination on time had there been no delay. Call that speed "s" in km/h.
Since 200 km is "halfway", the total distance must be 400 km.
time = distance / speed
total time = (time for first half) + (delay) + (time for second half)
400/s = 200/s + 1 + 200/(s+10) . . . .times are in hours, distances in km
200/s = 1 + 200/(s+10) . . . . . . . . . . subtract 200/s
200(s+10) = s(s+10) +200s . . . . . . .multiply by s(s+10)
0 = s² +10s - 2000 . . . . . . . . . . . . . .subtract the left side
(s+50)(s-40) = 0
Solutions are s = -50, s = 40
The speed of the bus before the traffic holdup was 40 km/h.
Answer:
x = 
Step-by-step explanation:
=> 3+2(2x+8) = 1
=> 3+4x+16 = 1
=> 4x+19 = 1
=> 4x = 1-19
=> 4x = -18
<u><em>Dividing both sides by 4</em></u>
=> x = 
=> x = 
Circumscribed circle for a triangle is circle where the circumference would cut all the vertices in the triangle
the diagram already has arcs marked from point Y, with the aim of drawing a perpendicular bisector to line XY.
Therefore the next step is to place the compass on point X and draw two arcs which cross the arcs already drawn (second option is the correct answer)
The points at which the 2 arcs meet are connected to each other making a perpendicular bisector that would cut XY perpendicularly.
Next step is to place the compass at Z and draw 2 arcs between Z and Y points, similarly place the compass at Y and draw 2 arcs that would cut the arcs already drawn.
Then connect the two points where the arcs meet and thats the perpendicular bisector of ZY.
The point at which perpendicular bisectors of XY and YZ meet is known as the circumcenter. Place the compass on this point with the radius any of the three points X,Y or Z and draw the circle
Answer:
2-square root 3, 2+ square root 3
Step-by-step explanation:
If the polynomial has rational and real coefficients, the roots will be "conjugates" of each other. That is, the sum of the root should be a rational number. So, the irrational parts will be opposites, while the rational parts remain the same.
2±√3 . . . can be the roots of p(x) when p(x) has rational real coefficients