In each pari, apply the Gram Schmidt process to the given subset S of the inner product space V to obtain an orthogonal basis for span(S'). Then normalize the vectors in this basis to obtain an orthonormal basis ri for span(S'). and compute the Fourier coefficients of the given vector relative to 0. Finally, use Theorem 6.5 to verify your result. (a) V - R: \ S - {(1.0, 1).(0. I. I). (1.3.3)}, and x = (1,1,2)
(b) V - R;! . S = {(1, 1, I), (0,1, I), (0,0,1)}, and x - (1,0,1)
(c) V - P2(/?) with the inner product (f(x),g(x)) - J* f(t)g(t)dt, S= {\..r..r2 }. and h(x) = 1+X
(d) V - span(.S'). where S = {(1,2,0), (1 -1,2,4*)}, and x = (3 + i,4i, 1)
(e) V - R1 . S - {(2. 1,-2,4), (-2,1,-5,5), (-1,3,7,11)}, and x = (-11.8.-4.18)
(f) V = R4 , S= {(1,-2, -1,3), (3,6,3, 1), (1,4,2,8)}, and x = (-1,2,1,1) / \ \, K, /i,

region b will make the inequalities y < 2x + 1 and y > -x - 2 true ⇒ answer a

step-by-step explanation:

* lets study the graph and the inequality to solve the question

- the inequality y > -x - 2

∵ the form of the inequality is y > mx + c , m is the slope of the line

  and c is the y-intercept (the line intersect y-axis at point (0 , c)

∴ c = -2 that means the line intersect the y-axis at point (0 , -2)

∵ the sign of the inequality is >

∴ we shad the part over the line

  region a and b

- the inequality y < 2x + 1

∵ the form of the inequality is y > mx + c , m is the slope of the line

  and c is the y-intercept (the line intersect y-axis at point (0 , c)

∴ c = 1 that means the line intersect the y-axis at point (0 , 1)

∵ the sign of the inequality is <

∴ we shad the part under the line

  region b and c

∵ the common region between the two lines is region b

∴ region b will make the inequalities y < 2x + 1 and y > -x - 2 true

answer: (yeees, you need a poet.)

//blooming serenity//

beautiful in the morning

mysterious at night

wonderous nature souring

with all it's mystical might

the flower that blooms

to the fox that looms

nocturnal creatures

with various features

a blooming flower soon in sight

as a new little hatchling fears to take flight  

afraid to leave it's calming nest,

but dear little bird, mother knows best

as sure as there are stars in the sky,

life is not about who, where what, when or why-

it's the journey that follows

adventures we swallow

of   nature's deep wonders,

natural disasters and storm's thunders.

blooming serenity

life and nature, truly is heavenly.

-kerstien lehman

step-by-step explanation:

let's first notice, the center of the hyperbola is at the origin, since the focus points are equidistantly 3 units away from the origin, that also means that the distance "c" from the center to either focus is 3.

it should also be notable that, the traverse axis for this hyperbola is vertical, or namely over the y-axis and therefore, the positive fraction will be the one with y-variable.

the traverse "a" component or distance from the center to either vertex is not 2, is more than 2, so we can safely put it at √5 ≈ 2.24.

b: y = 2x + 2

step-by-step explanation:

y = 2x + 1

we want a line that is parallel, so it will have the same slope

the line is in the form y= mx+b, so the slope is 2

we have a slope and the y intercept (0,2)

since x is 0 we know the y point which is the y intercept

we can use y= mx+b to write the equation of the line

m is the slope and b is the y intercept

y = 2x +2