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
Answer:
Please see the attached picture for the full solution.
*From the 4th line of the 1st image, you could also expand it using
(a +b)²= a² +2ab +b² and
(a -b)²= a² -2ab +b².
When squaring a fraction, square both the denominator and numerator.
➣(a/b)²= a²/b²
Step-by-step explanation:
y=75+50(36 hours)=1875
give me brainliest
Answer:
-8+5√2 and -8-5√2
Step-by-step explanation:
Given the expression x² + 16x + 14 = 0
USing the general formulas
x = -16±√16²-4(14)/2
x = -16±√256-56/2
x = -16±√200/2
x = -16±10√2/2
x = -8±5√2
Hence the required solutions are -8+5√2 and -8-5√2
Answer:
64
Step-by-step explanation:
I did 90-26 I hope that’s what you’re looking for :))
