Suppose m is a positive integer. is the set consisting of 0 and all polynomials with coefficients in f and with degree equal to m a subspace ofp(f)?

For this case we don't have any problems for the conditions 1) and 3), but we have a problem with condition 2) since is not satisfied.

We just need to find a counterexample to show that the statement is False. If we find two elements in the subset provided S, and we show that the sum is not in S, then we have the counter example.

Let's say that we have two elements so both elements are in S, and if we apply the condition for the addition closed we got:

And by definition of S, 2 is not in S so then since we can't satisfy the closed addition property then S can't be a subsapce

Step-by-step explanation:

For this case the answer would be no.

It can't be a subspace because we not satisfy the condition of closed under addition.

We need to remember that a subset U of V is a subspace of V if and only if U satisfies the following 3 conditions

1) Additive identity

2) Closed under addition

3) Cloases under scalar multiplication

Proof

For this case we don't have any problems for the conditions 1) and 3), but we have a problem with condition 2) since is not satisfied.

We just need to find a counterexample to show that the statement is False. If we find two elements in the subset provided S, and we show that the sum is not in S, then we have the counter example.

Let's say that we have two elements so both elements are in S, and if we apply the condition for the addition closed we got:

And by definition of S, 2 is not in S so then since we can't satisfy the closed addition property then S can't be a subsapce

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet

9) i can't see the angle so whatever it is just do this: cos(angle) x hypotenuse and then use that answer and they hypotenuse and plug it into this: (answer you just got)^2 + b^2 = (hypotenuse)^2. the first answer will be the answer of the adjacent side and the second answer after this equation will be the opposite side.

10) y = 14 x = √147 or 12.12435565

11) x = 4.8

12) x = 4.2

13) x = 9.3

14) pq: 13.2

qr : 17.6

angle p: 53º

15) angle c: 54.3º

angle a: 35.7º

16) h = 21040

v = 1841

step-by-step explanation:

10) 7/sin(30) = 14 = y

7^2 + b^2 = 14^2

49 + b^2 = 196

b^2 = 196 - 49

√b^2 = √147

x = √147

11) cos(70) x 14 = 4.788282007

x = 4.8

12) tan(25) x 9 = 4.196768923

x = 4.2

13) sin(31) x 18 = 9.270685348

x = 9.3

14) pq: sin(37) x 22 = 13.23993051

pq: 13.2

qr: cos(37) x 22 = 17.56998122

qr : 17.6

angle p: 180 - (37 + 90) = 53º

15) cos-1(7/12) = 54.31466529

angle c: 54.3º

angle a: 180 - (90 + 54.31466529) = 35.68533471

angle a: 35.7º

16) 4 x 5280 = 21120

cos(5) x 21120

h = 21039.63202

h = 21040

sin(5) x 21120

v = 1840.729287

v = 1841