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.
Line (a) is y = -3 and line (b) is y = -2x
a straight line is 180° so:
x+2/3x=180
multiply everything with 3 so that you'll looser the denominator:
3x+2x=540
5x=540
x=540:5
x=108
(2/3x= 72)
It proves The function f(x) is an arithmetic sequence.
