Lets count...
3a...thats 1 term
3ab...thats 2
7b....thats 3
-2c....movin on up to 4
-4d....and we end at 5
so this expression has 5 terms since none of the terms can be combined :)
Answer:

Step-by-step explanation:
Hello,
a and b are the zeros, we can say that

So we can say that

Now, we are looking for a polynomial where zeros are 2a+3b and 3a+2b
for instance we can write

and we can notice that
so
![(x-2a-3b)(x-3a-2b)=x^2-5(a+b)x+6[(a+b)2-2ab]+13ab\\= x^2-5(a+b)x+6(a+b)^2+ab](https://tex.z-dn.net/?f=%28x-2a-3b%29%28x-3a-2b%29%3Dx%5E2-5%28a%2Bb%29x%2B6%5B%28a%2Bb%292-2ab%5D%2B13ab%5C%5C%3D%20x%5E2-5%28a%2Bb%29x%2B6%28a%2Bb%29%5E2%2Bab)
it comes

multiply by 3

Answer: 3
This is the simplified way
Answer:
1 3/10
Step-by-step explanation:
3 7/10 - 2 2/5 =
3 7/10 - 2 4/10 =
1 3/10
Height = x (This is given)
Length = 5x
Width = 6x
Volume = length × width × height
The equation will be: (5x)(6x)(x) = 1920
simplify:
30x² × (x) = 1920
30x³ = 1920
Divide both sides by 30
x³= 64
x = ∛64
x= 4
If x = height, then, the height is 4cm
5x = length
20cm = length
6x = width
24cm = width