Answer: its the first one "opposite sides with equal length"
Answer:
12,7
Step-by-step explanation:
use pathogaras theorem
c^2=a^2+b^2
Let the numbers be m and n. Then "<span>five times the quotient of two numbers" would be 5m/n.
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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:
all work is shown and pictured
