Answer:
the number of times in a month the train must be used, so that the total monthly cost without the pass is the same as the total monthly cost with the pass, is b. 24 times
Step-by-step explanation:
in normal purchase, train ticket (A) = $2.00
using frequent pass,
frequent pass (P) = $18
train ticket using frequent pass (B) = $1.25
Now, let assume the number of times in a month the train must be used = M
so,
A x M = P + (B x M)
$2.00 x M = $18 + ($1.25 x M)
($2.00 x M) - ($1.25 x M) = $18
M x ($2.00 - $1.25) = $18
M = $18 : $0.75
M = 24
Thus, the number of times in a month the train must be used is 24 times
A) A = 20,000 [Principle] * (0.05 [Interest] * 4 [Number of years]) = $4,000 interest
b) A = 20,000 [Principle] * (0.05 [Interest] * 2 [Number of years]) = $2,000 interest
Multiply the first fraction by six and get : 2x - 9y = 24
So if this system has no solutions, then b would be nine.
Answer:
<em>one</em><em> </em><em>solution</em><em> </em>
<em>no</em><em> </em><em>solution</em><em> </em>
<em>infinity</em><em> </em><em>solutions</em><em> </em>
Step-by-step explanation:
A: If the graphs of the equations intersect, then there is one solution that is true for both equations.
B; If the graphs of the equations do not intersect (for example, if they are parallel), then there are no solutions that are true for both equations.
C; If the graphs of the equations are the same, then there are an infinite number of solutions that are true for both equations
YAnswer:
Step-by-step explanation:
