<h3>Given</h3>
4 hundreds flats; 5 tens rods; 2 ones cubes
<h3>Find</h3>
The number of hundreds flats in each of 2 equal piles
<h3>Solution</h3>
When 4 flats are divided into two equal groups, each group will have ...
... 2 flats
_____
You can imagine doing this the way a card dealer might: first put 1 flat in each of 2 piles, then do the same for the remaining 2 flats. Each pile will end up with 2 flats.
— — — — —
You will have a problem if you continue with the tens rods. There is an odd number of those, so one of them will have to be exchanged for 10 ones cubes.
Answer:
Step-by-step explanation:
We can use the FOIL method to expand two multiplied binomials. It states that 
. The FOIL method stands for First(first terms) outer(outer terms) inner(inner terms) last(last terms).
So, we can expand our binomials now!
Answer:
x³ - (√2)x² + 49x - 49√2
Step-by-step explanation:
If one root is -7i, another root must be 7i. You can't just have one root with i. The other roos is √2, so there are 3 roots.
x = -7i is one root,
(x + 7i) = 0 is the factor
x = 7i is one root
(x - 7i) = 0 is the factor
x = √2 is one root
(x - √2) = 0 is the factor
So the factors are...
(x + 7i)(x - 7i)(x - √2) = 0
Multiply these out to find the polynomial...
(x + 7i)(x - 7i) = x² + 7i - 7i - 49i²
Which simplifies to
x² - 49i² since i² = -1 , we have
x² - 49(-1)
x² + 49
Now we have...
(x² + 49)(x - √2) = 0
Now foil this out...
x²(x) - x²(-√2) + 49(x) + 49(-√2) = 0
x³ + (√2)x² + 49x - 49√2
