Answer:
Step-by-step explanation:
x² + x - 2 = x² - x +2x -2 = x(x - 1) + 2*(x-1) = (x-1)(x+2)
8x² + 4x = 4x ( 2x + 1)
3x² + 10x + 8 = 3x² + 6x + 4x + 4*2 =3x * (x + 2) + 4 * ( x + 2) = (x + 2)(3x + 4)
Answer:
3956
Step-by-step explanation:
92 * 43
92 * 40 = 3680
92 * 3 = 276
3680 + 276 = 3956
Answer:
896
Step-by-step explanation:
Let's talk first about how many 3 digit numbers there are. The first 3 digit number is 100 and the last is 999. So there are 999-100+1 numbers that are 3 digits long. That simplifies to 900.
Now let's find how many of those have a sum for the digits being 1, then 2 ? Then take that sum away from the 900 to see how many 3 digit numbers have the sum of their digits being more than 2.
3 digit numbers with sum of 1:
The first and only number is 100 since 1+0+0=1.
We can't include 010 or 001 because these aren't really three digits long.
3 digit numbers with sum of 2:
The first number is 101 since 1+0+1=2.
The second number is 110 since 1+1+0=2.
The third number is 200 since 2+0+0=2.
That's the last of those. We could only use 0,1, and 2 here.... Anything with a 3 in it would give us something larger than or equal to 3.
So there are 900-1-3 numbers who are 3 digits long and whose sum of digits is greater than 2.
This answer simplifies to 896.
