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.
The answer is the last one 7/8 x -3/4 =6
Answer:
1. 20
2. 23
3. 6
Step-by-step explanation:
We have that:
f(x) = 2x
g(x) = x² + 1
f(g(x)) is the composite function of f and g. So
f(g(x)) = f(x²-1) = 2(x²+1) = 2x² + 2
1. f(g(3))
f(g(x)) = 2x² - 2 = 2(3)² + 2 = 18 + 2 = 20
2. f(3)+g(4)
f(3) = 2(3) = 6
g(4) = 4² + 1 = 17
f(3) + g(4) = 6 + 17 = 23
3. f(5) - 2g(1)
f(5) = 2(5) = 10
g(1) = (1)² + 1 = 2
f(5) - 2g(1) = 10 - 2*2 = 10 - 4 = 6
Answer:
there whould be 1176 botone on all the pairs combined
Step-by-step explanation:
hope this will help
