Answer:
$0.90
Step-by-step explanation:
A 10-minute call exceeds the 3-minute initial period by 7 minutes. The initial period charge is 37¢. The additional minute charge is 7·9¢ = 63¢. Then the regular charge for that call is ...
37¢ +63¢ = 100¢ = $1.00
The 10% discount reduces the charge by ...
10% × $1.00 = $0.10
so the final cost of the 10-minute call is ...
$1.00 -0.10 = $0.90 . . . . . cost of the call
6x+3y=15
6*2+3y=15
12+3y=15
3y=3
y=1
6x+3y=15
6*5+3y=15
30+3y=15
3y=-15
y=-5
The answers is y=1 & -5
5n + 34 = -2(1 - 7n)
5n + 34 = -2 + 14n
5n = -36 + 14n
-9n = -36
9n = 36
n = 4
Answer:
3 hours
Step-by-step explanation:
speed of Bharat = 12kilometers / hour
speed of Ingrid=14 kilometers / hour
For this problem we'll be using formula relating time, distance and speed.i.e
Distance = speed x time
Suppose Bharat is riding at a distance of 'b' kilometers, therefore time taken by him will be:
Time = distance/ speed
Time= b/12 hours.
Also, Ingrid can ride the distance at this time with speed of 14km/hr
The distance would be,
Distance = speed x time
Distance = 14 x (b/12) => 14b/12
Distance= 7b/6
Let '
' be the distance that Bharat have covered after time 't'
therefore,
= 12 x t
Let '
' be the distance that Ingrid have covered after time 't'
therefore,
= 14 x t
In order to find the time when they are 78 kilometers apart, we will add
and
, because they are travelling in opposite direction creating distance between them.
So,
+
= 78
( 14 x t) + (12 x t) =78
14t + 12t =78
26t= 78
t= 78/26
t= 3hours.
thus, it will take 3 hours until they are 78 kilometers apart