(10^25)-7 is divisible by

a) 2 b)3 c)9 d) both 2 & 3

The answer is b)3.

If the last digit of a number is divisible by 2, than the number is divisible by 2 (for example, in 828, the last digit is 8 which is divisible by 2 (8/2 = 4), so 828 is also divisible by 2).

If the sum of all the digits of a number is divisible by 3, then the number is divisible by 3 (for example, in 396, the sum of the digits is 18 (3+9+6=18), so 396 is also divisible by 3). 

Similarly, if the sum of all the digits of a number is divisible by 9, then the number is divisible by 9.

Now, let's check it out for 10²⁵ - 7. Let's first take simple examples:
10² - 7 = 100 - 7 = 93 (The last digit is not divisible by 2. Also, the sum of the digits is 9+3=12 and it is divisible only by 3, but not by 9).
10³ - 7 = 1000 - 7 = 993 (The last digit is not divisible by 2. Also, the sum of the digits is 9+9+3=21 and it is divisible only by 3, but not by 9).
10⁴ - 7 = 10000- 7 = 993 (The last digit is not divisible by 2. Also, the sum of the digits is 9+9+9+3=30 and it is divisible only by 3, but not by 9).

10²⁵-7 = There are many zeros, but based on the previous examples, we can make a conclusion that the last digit is not divisible by 2. Also, the sum of the digitsit is divisible only by 3, but not by 9.

1. isolate the variable, in this case, d. subtract 35.5 on both sides of the equation.2. 52 subtracted by 35.5 is 16.53. therefore your answer would be d= 16.5this is the basics of pre algebra, isolating variable through adding, subtracting, multiplying, and dividing, if you have any other questions to ask just comment!