____ is a line that passes through exactly one point on a circle

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 please help me

docker41 [41]

Ur missing numbers are -2, 6, and 36

Since the center point is (-2,6), the numbers are reversed in the equation to: (x+2) or(x- -2) and (y-6)

The radius is 6 units so u square it to get 36

7 0

3 years ago

The inequality describes the range of monthly average temperatures T in degrees Fahrenheit at a certain location. Find an equiva

kolezko [41]

Answer:

Now if the high and low monthly average temperatures satisfy the inequality, then the , monthly averages are always within 22 degrees of 43°F.

Step-by-step explanation:

The inequality describes the range of monthly average temperatures T in degrees Fahrenheit at a certain location.

The inequality expression is given as:

$|T-43|\leq 22$

now this expression could also be expressed as:

$-22\leq T-43\leq22\\\\-22+43 \leq T \leq 22+43\\\\21\leq T\leq 65$

Now if the high and low monthly average temperatures satisfy the inequality, then the , monthly averages are always within 22 degrees of 43°F.

( As the difference is 22 degrees to the left and right)

5 0

3 years ago

7176 rounded to 1 significant figure

kati45 [8]

Answer:

7000

Step-by-step explanation:

8 0

3 years ago

Meredith is playing games at an arcade to earn tickets that she can exchange for a prize. She has 294 tickets from previous visi

grin007 [14]

First, plug in 448 for y:

448=7x+294

Secondly, subtract 294 from both sides to isolate x:

154=7x

Divide both sides by 7:

22=x

Refine:

x=22

The answer is D)22

8 0

3 years ago

Read 2 more answers