Answer:
This question is very big, you should shorten your question a bit so that I can answer it quickly, the rest is just that I did not understand the question a bit
Answer:
<u>Option </u><u>D</u> (y = 5/6x -12).
Step-by-step explanation:
Hey there!
The equation of the line which passes through the point (12,-2) is (y+2) = m2(x-12)………(i) [Using one point formula].
According to the question, the first line passes through point (12,6) and (0,-4).
So,
Therefore, the slope of the line is 5/6.
Now as per the condition of parallel lines, m1 =m2 = 5/6.
So, keeping the value of m2 in equation (i), we get;
(y+2) = 5/6(x-12)
or, y = 5/6x - 12.
Therefore, the required equation is y = 5/6 X - 12.
<u>Hope</u><u> </u><u>it </u><u>helps</u><u>!</u>
Answer:
0.18
Step-by-step explanation:
14x(8-10)=6
Distribute 14x
112x-80x=6
32x=6
6/32=0.18
x=0.18
Hope this helps
........and is right :p
1 meter = 3.28 ft
A = L * l = 2 * 3 = 6 square meter
To tranform it into square feet: 6 * 3.28 = 19.68 sq ft
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.