Answer:
Step-by-step explanation:
Order shouldn't make any difference.
3 C 4 = 4!/(3! 1!) = 4
Let's see if that is right.
1 2 3
1 2 4
1 3 4
2 3 4
Any other combination will produce a duplicate if order doesn't matter. For example
2 4 3 is the same thing as 2 3 4
Answer:
=3x+27y−6
Step-by-step explanation:
3(3x−(2−6y+2x)+3y)
=(3)(3x+−2+6y+−2x+3y)
=(3)(3x)+(3)(−2)+(3)(6y)+(3)(−2x)+(3)(3y)
=9x−6+18y−6x+9y
=3x+27y−6
Answer:
A= 3, B=-5, C=2
Step-by-step explanation:
Put them in order of leading power.
3x^2-5x+2
We have two numbers say a, b. And a system of equations.
It looks a bit intimidating but it's actually very easy.
Instead of term a + b in the second equation we can say it's zero because the first equation describes a + b as being equal to 0.
So,
The above equation is not true therefore numbers a and b do not exists in the real number domain.
No solution.
Hope this helps.
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