<h3>
Answer: 21p^3+5p?-10p-3p'-2</h3><h3>
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Step-by-step explanation:
Answer:
648
Step-by-step explanation:
Running this in Python, with the code as follows,
import math
cur_numbers = [0] * 3
num = 0
for i in range(100, 1000):
cur_numbers[2] = i % 10
i = math.floor(i/10)
cur_numbers[1] = i % 10
i = math.floor(i/10)
cur_numbers[0] = i % 10
if(len(set(cur_numbers)) == 3):
num += 1
print(cur_numbers)
print(num), we get 648 as our answer.
Another way to solve this is as follows:
There are 9 possibilities for the hundreds digit (1-9). Then, there are 10 possibilities for the tens digit, but we subtract 1 because it can't be the 1 same digit as the hundreds digit. For the ones digit, there are 10 possibilities, but we subtract 1 because it can't be the same as the hundreds digit and another 1 because it can't be the same as the tens digit. Multiplying these out, we have
9 possibilities for the hundreds digit x 9 possibilities for the tens digit x 8 possibilities for the ones digit = 648
The awnser is 52 because it’s going up by prime numbers
Is your question true or false?
2abc - 3ab = -30? if a =2, b = -3 and c= 4
plug in
2(2)(-3)(4) - 3(2)(-3) = -30
Simplify
-48 + 18 = -30
-30 = -30
Both sides are equal so
It's true.
