Answer:
It is not a subspace.
Step-by-step explanation:
So, the polynomial degree of at most 3 is given below as;
V = { p (z) = b0 + b1z + b2z^ + ... + bnz^n. |n is less than or equal to 3 and b0, b1, b2,... are integers.
To determine whether a subset is a subspace of Pn, we have to check for the properties below;
(1). Zero vector property : that is, when polynomial, p(z) = 0 and 0 is an integer.
(2). Addition property= here, we have; p(z) + h(z) = (b0 + b1z + b2z^2 +....+ bnz^n) + ( c0 + c1z + c^2z^2 +... + cnz^n). That is the sum of integers.
(3). Scaler multiplication property: the coefficient here may not be real numbers therefore, the condition is not followed here.
Therefore, it is not a subspace of Pn.
STEP
1
:
y
Simplify —
3
Equation at the end of step
1
:
y
(((18•(x5))•(y3))+((6•(x2))•(y4)))+((((24•(x6))•—)•x)•y)
3
STEP
2
:
Equation at the end of step
2
:
y
(((18•(x5))•(y3))+((6•(x2))•(y4)))+((((23•3x6)•—)•x)•y)
3
STEP
3
:
Canceling Out:
3.1 Canceling out 3 as it appears on both sides of the fraction line
Equation at the end of step
3
:
(((18•(x5))•(y3))+((6•(x2))•(y4)))+((8x6y•x)•y)
STEP
4
:
Equation at the end of step
4
:
(((18•(x5))•(y3))+((2•3x2)•y4))+8x7y2
STEP
5
:
Equation at the end of step
5
:
(((2•32x5) • y3) + (2•3x2y4)) + 8x7y2
STEP
6
:
STEP
7
:
Pulling out like terms
7.1 Pull out like factors :
8x7y2 + 18x5y3 + 6x2y4 = 2x2y2 • (4x5 + 9x3y + 3y2)
Trying to factor a multi variable polynomial :
7.2 Factoring 4x5 + 9x3y + 3y2
Try to factor this multi-variable trinomial using trial and error
Factorization fails
Final result :
2x2y2 • (4x5 + 9x3y + 3y2)
Answer:54
Step-by-step explanation: You simply minus $66 from $120. I am fairly sure that i am right. If my answer is wrong, then please forgive me.
Answer:
2.5
Step-by-step explanation:
2.5 is the constant of proportionality
