There are 10,000 total four-digit numbers (1000 through 9999).
Multiples of 2 end in 0, 2, 4, 6, and 8. There are 9*10*10*5 = 4500 four-digit multiples of 2.
Multiples of 5 end in 0 or 5. There are 9*10*10*2 = 1800 four-digit multiples of 5.
There is redundancy between the two sets of numbers, namely those that end in 0, which are both multiples of 2 and 5. There are 9*10*10*1 = 900 four-digit multiples of both 2 and 5.
Then there are 4500 + 1800 - 900 = 5400 total four-digit numbers that are either multiples of 2 or 5, which means the remaining 4600 numbers are neither multiples of 2 nor 5.
Answer:
1 2/5
Step-by-step explanation:
Step 1:
1.4 = 140/100
Step 2:
140/100 = 70/50
Step 3:
70/50 = 35/25
Step 4:
35/25 = 7/5
Step 5:
7/5 = 1 2/5
Answer:
1 2/5
Hope This Helps :)
1/2DF = EF
1/2(2y) = 8y - 3
y = 8y - 3
- Subtract 8y from both sides.
-7y = -3
y = 3/7
- Plug 3/7 for y into DF and EF.
DF: 2(3/7) = 6/7
EF: 8(3/7) - 3 = 24/7 - 3 = 3/7
EF = DE because they are bisected, so DE = 3/7
