It is instructed to subtract (2a−3b+4c) from the sum of (a+3b−4c),(4a−b+9c) and (−2b+3c−a).
So, First we will do the sum of the three given polynomials,
Sum =(a+3b−4c)+(4a−b+9c)+(−2b+3c−a)
=(a+4a−a)+(3b−b−2b)+(−4c+9c+3c)
=4a+8c
Now, we can perform the subtraction,
∴ Required difference
=(4a+8c)−(2a−3b+4c)
=4a+8c−2a+3b−4c
=2a+3b+4c
Answer:
The value of A is also 40
Step-by-step explanation:
My reasoning why is because its the same for each.
It is already IN standard form. If you don't want it in standard form notation it would be 74,700
The statement is correct.
3/4 divided by 1/2 can be written as 3/4 x 2/1, because even dividing fractions you can flip the second one and multiply instead.
This gives 6/4, which can be written as 3/2. This is top heavy, meaning it is greater than 1. It could also be said as 3 divided by 2, and half of 3 is 1.5, greater than 1.
For the second one, 1/2 divided by 3/4 can be written as 1/2 x 4/3, which is 4/6, also written as 2/3.
2/3 is less than 3/3, making it less than 1.
This means Pilar is correct for both cases as the former is indeed greater than 1 and the latter is less than.
