Number 6 would be
because all of the other numbers are irrational numbers
Answer:Your left hand side evaluates to:
m+(−1)mn+(−1)m+(−1)mnp
and your right hand side evaluates to:
m+(−1)mn+(−1)m+np
After eliminating the common terms:
m+(−1)mn from both sides, we are left with showing:
(−1)m+(−1)mnp=(−1)m+np
If p=0, both sides are clearly equal, so assume p≠0, and we can (by cancellation) simply prove:
(−1)(−1)mn=(−1)n.
It should be clear that if m is even, we have equality (both sides are (−1)n), so we are down to the case where m is odd. In this case:
(−1)(−1)mn=(−1)−n=1(−1)n
Multiplying both sides by (−1)n then yields:
1=(−1)2n=[(−1)n]2 which is always true, no matter what n is
The product of something means multiplying the terms together.
(2x+3) (4x^2-5x+6)
Secondly you need to distribute the terms to each other (Think of problems like FOIL)
2x * 4x^2 + 2x(-5x) + 2x * 6 + 3 * 4x^2 + 3(-5x) + 3 * 6
Then you must take into account that some of the numbers are negative. (minus-plus rules!)
2x * 4x^2 - 2x * 5x + 2x * 6 + 3 * 4x^2 - 3 * 5x + 3 * 6
Now is the tricky part of simplifying everything.
2x * 4x^2 = 8x^3
2x * 5x = 10x^2
2x * 6 = 12x
3 * 4x^2 = 12x^2
3 * 5x = 15x
3 * 6 = 18
8x^3 - 10x^2 + 12x + 12x^2 - 15x + 18
Then you group like terms.
8x^3 - 10x^2 + 12x^2 - 3x + 18
8x^2 + 2x^2 - 3x + 18
The trickiest part of this is distributing all of the terms within the parentheses, once you've done that, it's smooth sailing!
Answer:
Honestly agreed, i feel like the smaller grades k- 6th grade are useful. where yoh learn to add, multiply, read time and such.
7 books with a cost of 28$ then a 2$ charge which will put him right at 30
