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sineoko [7]
3 years ago
14

The arcs in the diagram are labeled with their measures in degrees. What is the measure of arc uv?

Mathematics
1 answer:
IrinaVladis [17]3 years ago
3 0

Answer:

10 steps

Step-by-step explanation:

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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_2is 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
Pearson realize help please!
yulyashka [42]

Answer:

I think it would be 15

Step-by-step explanation:

Because if you split the number line into 6 different sections (with the 15 on the left side and the 16 on the right side) and shade/fill in 2 of those, you would be closer to 15.

6 0
3 years ago
5. Find the circumcenter of the triangle. XXX CU 3 (4,2) KNNYA XHENNARO 3 ****************++++++ 5 NET Virtual cla 44,-3) WHEREw
Elan Coil [88]

The circumcenter of a right angled triangle is midpoint of the hypotenuse side.

Determine the midpoint of the hypotenuse.

\begin{gathered} x=\frac{-4+0}{2} \\ =\frac{-4}{2} \\ =-2 \end{gathered}

For y coordinate,

\begin{gathered} y=\frac{2-3}{2} \\ =\frac{-1}{2} \end{gathered}

So, circumcenter of triangle is,

(-2,\frac{-1}{2})

8 0
1 year ago
A farn stand sells apple pies
inessss [21]
I need more information other than “a farn stand sells apple pies” to help you.
7 0
3 years ago
2.8962 to the nearest hundredth do not write extra zeros?<br>​
Alexxx [7]

Answer:

2.9

Step-by-step explanation:

Plz mark as brainliest!!!

5 0
3 years ago
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