If 5 consecutive integers is 205,
then a + b + c + d + e = 205
but also, each integer is separated by a difference of 1
⇒ a + (a + 1) + (a + 2) + (a + 3) + (a + 4) = 205
⇒ 5a + 10 = 205
⇒ 5a = 195
⇒ a = 39
∴ third term = 39 + 2
= 41
5men because half of them would probably not make it
35/8 = (4 * 8 = 32) and 3 left over so answer is 4 3/8
Answer:
Randy has eight $5 bills and nine $1 bills
Step-by-step explanation:
Randy needs $50.00
And we know that he his only $1.00 short, so he has $49.00
let's define:
x = number of $1 bills that he has
y = number of $5 bills that he has.
then:
x*$1 + y*$5 = $49
We know that he has one more $1 bills than $5 bills.
we can write this as
x = y + 1
So we have a system of two equations and two variables:
x*$1 + y*$5 = $49
x = y + 1
First we can see that the variable "x" is isolated in the second equation, now we can replace that in the other equation:
x*$1 + y*$5 = $49
(y + 1)*$1 + y*$5 = $49
now we can solve this for y.
y*$1 + $1 + y*$5 = $49
y*($1 + $5) = $49 - $1 = $48
y*$6 = $48
y = $48/$6 = 8
He has 8 $5 bills
and we know that:
x = y + 1
x = 8 + 1 = 9
he has 9 $1 bills.
