Answer:
21 quarters and 19 dimes
Step-by-step explanation:
Create a system of equations where q is the number of quarters and d is the number of dimes.
0.25q + 0.1d = 7.15
q + d = 40
Solve by elimination by multiplying the bottom equation by -0.25
0.25q + 0.1d = 7.15
-0.25q - 0.25d = -10
Add them together and solve for d:
-0.15d = -2.85
d = 19
Then, plug in 19 as d into the second equation to solve for q:
q + d = 40
q + 19 = 40
q = 21
So, there are 21 quarters and 19 dimes
9514 1404 393
Answer:
BC = 35
Step-by-step explanation:
The sum of the segments is the overall length.
AB +BC = AC
(2x +6) + (6x +5) = 51 . . . . substitute given values
8x +11 = 51 . . . . . . . . . . . . . collect terms
8x = 40 . . . . . . . subtract 11
x = 5 . . . . . . . . . divide by 8
BC = 6x+5 = 6(5) +5 . . . . . . . . use the value of x to find BC
BC = 35
For this case, the first thing we must do is define variables.
We have then:
n: number of cans that each student must bring
We know that:
The teacher will bring 5 cans
There are 20 students in the class
At least 105 cans must be brought, but no more than 205 cans
Therefore the inequation of the problem is given by:
Answer:
105 <u><</u> 20n + 5 <u><</u> 205
the possible numbers n of cans that each student should bring in is:
105 <u><</u> 20n + 5 <u><</u> 205
Answer:
distance = 585600 m
Step-by-step explanation:
change 2 hours 20 minutes to seconds
2 hrs * 60 minutes/1hr * 60 sec/min = 7200 sec
20 minutes * 60 sec/min = 120 sec
2 hr 20 min = 7320 seconds
distance = rate * time
distance = 80 m/s * 7320 s = 585600 m
73.14 $ knowing that the tax would be 4.14$, added to 69$ would be 73.14$
