103^2-102^2
The property that applies here is a^2-b^2 = (a+b)(a-b)
so the answer is 103^2-102^2=(102+103)*(103-102)= numerically it is 205( as 103+102 = 205 and 103-102 = 1.....so 205*1 = 205)
The thing is that usually the factorization ends at (102+103)(102-103) ......
Answer:
2/9, because it is 2 divided by 9. i think this is correct
Step-by-step explanation:
Answer:
Step-by-step explanation:
Factor out x.
Simplify.
(+/-) <span>1 x (+/-) 232 = 232
</span>(+/-) <span>2 x (+/-)116 = 232
</span>(+/-) 4 x (+/-) <span>58 = 232
</span>(+/-) 8 x (+/-) <span>29 = 232
</span>(+/-) 29 x (+/-) <span>8 = 232
</span>(+/-) 58 x (+/-) <span>4 = 232
</span>(+/-) 116 x (+/-) <span>2 = 232
</span>(+/-) 232 x (+/-) 1 = 232
So the question if find y'' for 4y=cos2x
First, find y by dividing by 4. y = cos2x/4
Then, take the first derivative. y' = -sin2x/2
Then, take the second dervative. y'' = -cos2x