Pls help this should be easy for u guys

Which of the following sets are subspaces of R3 ?

Ratling [72]

Answer:

The following are the solution to the given points:

Step-by-step explanation:

for point A:

$\to A={(x,y,z)|3x+8y-5z=2} \\\\\to for(x_1, y_1, z_1),(x_2, y_2, z_2) \varepsilon A\\\\ a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=3(aX_l +bX_2) + 8(ay_1 + by_2) — 5(az_1+bz_2)\\\\=a(3X_l+8y_1- 5z_1)+b (3X_2+8y_2—5z_2)\\\\=2(a+b)$

The set A is not part of the subspace $R^3$

for point B:

$\to B={(x,y,z)|-4x-9y+7z=0}\\\\\to for(x_1,y_1,z_1),(x_2, y_2, z_2) \varepsilon B \\\\\to a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=-4(aX_l +bX_2) -9(ay_1 + by_2) +7(az_1+bz_2)\\\\=a(-4X_l-9y_1+7z_1)+b (-4X_2-9y_2+7z_2)\\\\=0$

$\to a(x_1,y_1,z_1)+b(x_2, y_2, z_2) \varepsilon B$

The set B is part of the subspace $R^3$

for point C: $\to C={(x,y,z)|x$

In this, the scalar multiplication can't behold

$\to for (-2,-1,2) \varepsilon C$

$\to -1(-2,-1,2)= (2,1,-1)$ ∉ C

this inequality is not hold

The set C is not a part of the subspace $R^3$

for point D:

$\to D={(-4,y,z)|\ y,\ z \ arbitrary \ numbers)$

The scalar multiplication s is not to hold

$\to for (-4, 1,2)\varepsilon D\\\\\to -1(-4,1,2) = (4,-1,-2)$ ∉ D

this is an inequality, which is not hold

The set D is not part of the subspace $R^3$

For point E:

$\to E= {(x,0,0)}|x \ is \ arbitrary) \\\\\to for (x_1,0 ,0) ,(x_{2},0 ,0) \varepsilon E \\\\\to a(x_1,0,0) +b(x_{2},0,0)= (ax_1+bx_2,0,0)\\$

The $x_1, x_2$ is the arbitrary, in which $ax_1+bx_2$ is arbitrary

$\to a(x_1,0,0)+b(x_2,0,0) \varepsilon E$

The set E is the part of the subspace $R^3$

For point F:

$\to F= {(-2x,-3x,-8x)}|x \ is \ arbitrary) \\\\\to for (-2x_1,-3x_1,-8x_1),(-2x_2,-3x_2,-8x_2)\varepsilon F \\\\\to a(-2x_1,-3x_1,-8x_1) +b(-2x_1,-3x_1,-8x_1)= (-2(ax_1+bx_2),-3(ax_1+bx_2),-8(ax_1+bx_2))$

The $x_1, x_2$ arbitrary so, they have $ax_1+bx_2$ as the arbitrary $\to a(-2x_1,-3x_1,-8x_1)+b(-2x_2,-3x_2,-8x_2) \varepsilon F$

The set F is the subspace of $R^3$

5 0

3 years ago

Please help me need help me please please help me

Reil [10]

9514 1404 393

Answer:

(3) y = (x -3)²

Step-by-step explanation:

Try the first value of x in each function:

(1) y = 3² -3 ≠ 0

(2) y = (3 +3)² ≠ 0

(3) y = (3 -3)² = 0 . . . . . . matches the table value for f(3)

(4) y = 3² +3 ≠ 0

__

The only viable choice is (3).

5 0

3 years ago

Jane was paid $316 for 40 hours of work. Use compatible numbers to find about how much

Vinvika [58]

Answer:

$7.9 /hour

Step-by-step explanation:

$316/40 = $7. 9per hour

4 0

3 years ago

A rectangular deck is 12 ft by 14 ft. When the length and width are increased by the same amount, the area becomes 288 sq . By h

julia-pushkina [17]

Well to get the area of 288 the dimensions would be a couple of things but 16 ft by 18 ft seems to work, so the dimensions increase by 4 ft on each side.

12+4=16
14+4=18

18*16=288

6 0

3 years ago

Read 2 more answers

An independent-measures research study uses a total of 18 participants to compare two treatment conditions. If the results are u

Luba_88 [7]

Answer:

The answer is false

Step-by-step explanation:

In a sample above 30 obs like this the confidence interval is defined as

X+- t* (s/sqrt(n)) where X is the mean t the tvalue for a given confidence level, n the size of sample and s standar deviation.

To find de appropiate value of t we must see the T table where rows are degrees of freedom and columns significance level

The significance is obtained:

significance = 1 - confidence level = 1 - 0.9 = 0.10

Degrees of freedom (df) for the inteval are

df = n - 1 = 18 - 1 = 17

So we must look for the value of a t with 17 values and significance of 0.10 which in t table is 1.740 not 1.746 ( thats the t for 16 df)

6 0

3 years ago