Radius = 5 cm
Height = 8 cm
Volume = πr²h
Volume = π(5 cm)² × 8 cm
Volume = π × 25 cm² × 8 cm
Volume = 200π cm³
$20 minus $15 = $5 difference
$5 divided by $0.20 = 25 checks
*Now the accounts are even, except I haven't done the checks for the second account.*
25 checks times $0.10 = $2.50
$2.50 divided by $0.20 = 12.5 checks (round up in this case to 13)
*Now, you need to do 13 more checks on the second account*
13 checks times $0.10 = $1.30
$1.30 divided by $0.20 = 6.5 checks (round up in this case to 7)
*Now, you need to do 7 more checks on the second account*
7 checks times $0.10 = $0.70
$0.70 divided by $0.20 = 3.5 checks (round up in this case to 4)
*Now, you need to do 7 more checks on the second account*
4 checks times $0.10 = $0.40
$0.40 divided by $0.20 = 2 checks
*Now, you need to do 2 more checks on the second account*
2 checks times $0.10 = $0.20
$0.20 divided by $0.20 = 1 check
*Now, you need to do 1 more checks on the second account*
1 checks times $0.10 = $0.10
$0.10 divided by $0.20 = 0.5 checks (round up in this case to 1)
*Now, you need to do 1 more checks on the second account*
1 checks times $0.10 = $0.10
$0.10 divided by $0.20 = 0.5 checks
25 + 13 + 7 + 4 + 2 + 1 + 1 = 53 checks
Check your work!
Account #1- $15 + (53 times $0.20) = $25.60
Account #2- $20 + (53 times $0.10) = $25.30
Answer
53 checks
34=2x + 16
18=2x
x=9
Robert used 9 grams of iron filings in his first experiment.
(X+1) y (2x-1) this is the answer
Answer:
5040,56
Step-by-step explanation:
We have to construct pass words of 4 digits
a) None of the digits can be repeated
We have total digits as 0 to 9.
4 digits can be selected form these 10 in 10P4 ways (since order matters in numbers)
No of passwords = 10P4
= 
b) start with 5 and end in even digit
Here we restrain the choices by putting conditions
I digit is compulsorily 5 and hence only one way
Last digit can be any one of 0,2,4,6,8 hence 5 ways
Once first and last selected remaining 2 digits can be selected from remaining 8 digits in 8P2 ways (order counts here)
=56
