algebraic word problem (10 points)?
ok im studying for my finals and im having a hard time answering this word problem below
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" A bank teller has $10,000 in $20 bills and $50 bills. If the teller has a total of 320 bills, how many of each kind are there."
i know i have to set up the word problem in some type of algebraic expression but i dont how to
please answer the word problem and show me what you did..step by step.
best answer = 10 points!
Comments
let x be the number of 20 bills
let y be the number of 50 bills
Since you do not know the number of each bill that the teller has.Then
20x + 50y = 10000.....eqn 1
Also the teller have a total of 320 bills
then x + y = 320 (Total number of bills)...eqn2
These are simultaneous eqns.
Using the sustitution method
Solve for x in eqn.2
x = 320 - y .......eqn 3
Then substitute eqn 3 in eqn 1
20 (320 - y) + 50y =10000
6400 - 20y +50y =10000
6400 + 30y = 10000
30y = 3600
y = 120
therefore x = 320 - 120 = 200
Therefore there are 200 $20 bills and 120 $50 bills
1. x+y=320
where x= #of 20's and y=#of 50's.
2. therefore; (x(20)) + (y(50))=10,000
substitute y in equation 2 for 320-x in equation 1 and then solve for x.
(x(20))+ ((320-x)(50))=10,000
once solving for x(the number of 20 dollar bills) then insert number into equation 1. and solve for y.
Answer: to check from your work......x=200 and y=120
the portion of a circle makes use of the equation : ?(r^2) - pie x radius squared which finally leads to - ? (p - 5)^2 (p - 5)^2 = p^2 - 10p + 25 putting it altogether: ? (p^2 - 10p + 25) your final answer desire that helps