I’m guessing that this is for systems of equations?
n - nickels
d - dimes
These are the equations you start off with.
n = 5d-2
0.95 = 0.10d+0.05n
Substitute the top equation for n into the variable n in the bottom equation.
0.95 = 0.10d+0.05(5d-2)
Solve for d.
0.95 = 0.10d+0.25d-0.10
1.05 = 0.35d
3 = d
Substitute d into the top equation and solve for n.
n = 5(3)-2
n = 15-3
n = 12
There are 3 dimes and 12 nickels in the coin purse! Hope this helped <3
The next time they wil leave together is 6:00. The LCM of 8, 12, and 10 is 120. So 120/6=2, so two hours. 6:00.
The answer is 7 you have to add 19 and 15 together and when you get the answer subtract that from 41 and you get 7
Answer:
• Group the h terms by organised term arrangement :
• Then using distributive property, factorise out the value h so that the reverse is true.
• for the variable "lw", divide it by h in order to add it to the bracket of (w + l). Make sure the reverse is true:
• finally, completely factorise out the value h
