Million has 6 zeoes
34,000,000+256,00*400
PEMDAS
multiply first
256,000*400=102,400,000
add to 34,000,000
136,400,000
now
must be in form
(x)(10^m)
such taht
1<x<10
and x times 10^m is the original number
1.364*10^8
Answer: 1513.50=961.50+46x
The number of players the team can bring to the tournament = 12.
Step-by-step explanation:
Let x= the number of players the team can bring to the tournament.
Total money raised = Rent of bus + (Cost per player) (Number of players)
Hence, the number of players the team can bring to the tournament = 12.
Answer:
26
Step-by-step explanation:
Problem: f(x) = 28 + x + 7
Plug in f(-9)
New equation: f(-9) = 28 + (-9) + 7
Simplify: f(-9) = 26
<u>Answer: f(-9) = 26</u>
9514 1404 393
Answer:
250
Step-by-step explanation:
Let 'a' represent the number of adult tickets.
a +(a -73) = 427
2a = 500 . . . . . add 73
a = 250 . . . . . . divide by 2
250 adult tickets were sold.
_____
<em>Additional comment</em>
I call this a "sum and difference problem" because we are given the total of two values and the difference between them. As you can see here, the larger of the two values is the average of the given numbers, their sum divided by 2. This is the generic solution to such a problem: the larger number is the average of the given sum and difference.
The nonpermissible replacement is a value that make the denominator equal to zero
