Question

-Du. The taxi fare in city is as follows
For the first kilometre, the
өне Шу -
30 9*1 Дие -
quent distant it is Rs. 5 pen km.
Taking the distance covered
I as a km and total fare as Rs. y
write alineas equation for this
info. and draw it's graph.​

Answer 1

Answer:

if it is the question frm class 9 maths ex 4.3 Q. 4 my answer is this

Step-by-step explanation:

• total distance travelled= x (km)
• fare of 1st one km= rupees 8
• fare of per/km after the 1st km= rupees 5 (after u travelled one km this is the distance u travelled after that)
• so, fare of the remaining distance= (x-1)5 because x is the total distance n u hv already traveled one km n ur rupees 8 is confirm to give. Therefore if u subtract ur one km frm the total distance, u will get the remaining distance n for per/km u travel is rupees 5 (like just for eg,, ur total distance is n u hv travelled one km so u will simply minus 1 frm 9 n u will get 8 which is the remaining distance u traveled n multiply it by 5 as it is the fare of per/km u travelled )
• total fare= (x-1)5 + 8= y

= 5x- 5 +8 = y

= 5x - y +3= y (ur equation)

x=0

5 × 0 - y + 3 = 0

- y = -3

y = 3

x= 1

5 × 1 - y + 3 = 0

5 - y + 3 = 0

8 = y

so, we have two points (0,3) and (1,8)

• just plot these points on the graph n join the points name the line 5x - y + 3 = y
• N UR DONE!!!!

HOPE IT HELPS YOU