First you have to subtract 1.3 from 0.75 which equals -0.55. Then you have to add 3/7 to 2 which equals 2 and 3/7. Then you divide 2 and 3/7 by -0.55 which equals 0.9785 and that’s your answer. Sorry I don’t know the answer for the second problem but I hope this helped.
Answer:
I think 24 will be ur ans_______
Answer:
16 =x
Step-by-step explanation:
The angles are corresponding angles and since the lines are parallel, they are equal
5x+18
=7x-14
Subtract 5x from each side
5x+18 -5x=7x-5x-14
18 = 2x-14
Add 14 to each side
18+14 = 2x-14+14
32 = 2x
Divide by 2
32/2 = 2x/2
16 =x
Answer:
a) 6 mins
b) 70km/h
c) t= 45
Step-by-step explanation:
a) The bus stops from t=10 to t=16 minutes since the distance the busvtravelled remained constant at 15km
Duration
= 16 -10
= 6 minutes
b) Average speed
= total distance ÷ total time
Total time
= 24min
= (24÷60) hr
= 0.4 h
Average speed
= 28 ÷0.4
= 70 km/h
c) Average speed= total distance/ total time
Average speed
= 80km/h
= (80÷60) km/min
= 1⅓ km/min
1⅓= 28 ÷(t -24)
<em>since</em><em> </em><em>duration</em><em> </em><em>for</em><em> </em><em>return</em><em> </em><em>journey</em><em> </em><em>is</em><em> </em><em>from</em><em> </em><em>t</em><em>=</em><em>2</em><em>4</em><em> </em><em>mins</em><em> </em><em>to</em><em> </em><em>t</em><em> </em><em>mins</em><em>.</em>
(t -24)= 28
t - 32= 28
t= 32 +28
t= 60
t=
t= 45
*Here, I assume that this is a displacement- time graph, so the distance shown is the distance of the bus from the starting point because technically if it is a distance-time graph, the distance would still increase as the bus travels the 'return journey'.
Thus, distance is decreasing after t=24 and reaches zero at time= t mins so that is the return journey. (because when the bus returns back to starting point, displacement/ distance from starting point= 0km)
Answer:
try hard
Step-by-step explanation: