To cover a distance of 240 km has to increase a speed of 5 per hour to reach in time time. If the time Lost was 40 minutes during traffic find the original speed

The initial speed is approximately 104.3 km

Step-by-step explanation:

Let Distance = 240 km

Time = ?

Speed =

Since

Speed = Distance/Time

Distance = Speed × Time

When arrival is late:

240 = S×T

When arrival is on time:

240 = (S+5)×T2

But 40minutes = 2/3 hours

So,

T2 = T - 2/3

240 = (S+5)×(T - 2/3)

240 = ST - 2S/3 + 5T - 10/3

ST - 2S/3 + 5T = 240 + 10/3

S(T - 2/3) + 5T = 730/3

S(T - 2/3) = 730/3 - 5T

S = (730/3 - 5T)/(T - 2/3)

But 240 = ST

S = 240/T

240/T = (730/3 - 5T)/(T - 2/3)

(T - 2/3)/T = (730/3 - 5T)/240

1 - 2/3T = 730/720 - 5T/240

1 + 73/72 = (2/3 + 5/24)T

145/72 = (7/8)T

T = 145/72 × 8/7

= 1160/504

T = 145/63

S = 240/S

= 240×63/145

= 3024/29

≈ 104.3km

10a^3+277a^2-132a-792

step-by-step explanation:

(2a+11)(5a-12)(foil)

=10a^2-24a+55a-132 =10a^2+31a-132

(10a^2+31a-132)(a+6)

=10a^3+60a^2+31a^2+186a^2-132a-792 =10a^3+277a^2-132a-792

i hope this is what you are looking for : )