34. [M] Let B D f1; cost; cos2 t; : : : ; cos6 tg and C D f1; cost; cos 2t; : : : ; cos 6tg. Assume the following trigonometric identities (see Exercise 37 in Section 4.1). cos 2t D 1 C 2 cos2 t cos 3t D 3 cost C 4 cos3 t cos 4t D 1 8 cos2 t C 8 cos4 t cos 5t D 5 cost 20 cos3 t C 16 cos5 t cos 6t D 1 C 18 cos2 t 48 cos4 t C 32 cos6 t Let H be the subspace of functions spanned by the functions in B. Then B is a basis for H, by Exercise 38 in Section 4.3. a. Write the B-coordinate vectors of the vectors in C, and use them to show that C is a linearly independent set in H. b. Explain why C is a

the simplified form is

step-by-step explanation:

we want to simplify the expression,

let us change the middle bar to a normal division sign to obtain,

we multiply the first fraction by the reciprocal of the second fraction to obtain,

we factor to obtain,

we cancel the common factors to get,

we simplify to get,

the correct answer is b

(1, –1), (16, –4), (36, –6)

step-by-step explanation:

given that the function is

this can also be written as

we have to graph this function and check which list of points lie on the graph

see attachment for graph:

we find that the points lying are (1,-1) (16,-4) and (36,-6)

since square root is there we find that x can only be posiitve but y can be since square root both positive and negative.

35x 3= 105

step-by-step explanation:

35+35= 70   70 + 35= 105