Answer:
Number of positive four-digit integers which are multiples of 5 and less than 4,000 = 600
Explanation:
Lowest four digit positive integer = 1000
Highest four digit positive integer less than 4000 = 3999
We know that multiples of 5 end with 0 or 5 in their last digit.
So, lowest four digit positive integer which is a multiple of 5 = 1000
Highest four digit positive integer less than 4000 which is a multiple of 5 = 3995.
So, the numbers goes like,
1000, 1005, 1010 .....................................................3990, 3995
These numbers are in arithmetic progression, so we have first term = 1000 and common difference = 5 and nth term(An) = 3995, we need to find n.
An = a + (n-1)d
3995 = 1000 + (n-1)x 5
(n-1) x 5 = 2995
(n-1) = 599
n = 600
So, number of positive four-digit integers which are multiples of 5 and less than 4,000 = 600
Answer:
625
Step-by-step explanation:
All it is saying to do is to do -5x-5x-5x-5
Answer D zero there no solution
Answer:
37 and 38
Step-by-step explanation:
let the consecutive integers be x-1 and x
If their sum is 75
x-1+x = 75
2x - 1 = 75
2x = 75+1
2x = 76
x = 76/2
x = 38
First integer = x - 1
First integer = 38-1 = 37
Second integer is 38
Hence the integers are 37 and 38
Answer:
No solution
Step-by-step explanation:
Actually, you have only one equation here. We will solve it for x.
Performing the indicated multiplication, we obtain:
10-2(2x+1)=4(x-2) => 10 - 4x - 2 = 4x - 8, or 10 - 2 = -8. This can never be true, so this equation has no solution.
