Answer:
Step-by-step explanation:
Let +s+ = the speed of the passenger train
+s+-+20%5D+ = the speed of the freight train
Let +t+ = time in hours for both trains
-------------------------------------
Equation forpassenger train:
(1) +280+=+s%2At+
Equation for freight train:
(2) +180+=+%28+s-20+%29%2At+
------------------------
(1) +t+=+280%2Fs+
Substitute (1) into (2)
(2) +180+=+%28+s-20+%29%2A%28+280%2Fs+%29+
(2) +180s+=+280s+-+5600+
(2) +100s+=+5600+
(2) +s+=+56+
and
+s+-+20+=+36+
The speed of the passenger train is 56 mi/hr
The speed of the freight train is 36 mi/hr
-------------------------------------
check:
(1) +280+=+56%2At+
(1) +t+=+5+
and
(2) +180+=+%28+56-20+%29%2At+
(2) +180+=+36t+
(2) +t+=+5+
OK
Answer:
maybe 6.24
sorry if wrong
Step-by-step explanation:
Answer:
let number of bags of popcorn = b
let number of drinks = d
at most = maximum
b + d ≤ 12
First let us find the length of JN
JN =JK +KN
JN = 82+105 = 187
JN=187..............(1)
UC= JN -( JH +HU+CN)
We are given :
JH= 22, HU = 96 and CN = 51
plugging all the values we get
UC = 187-( 22+96+51)
UC =187 -169
UC = 18
Answer is UC =18 ( option B)
