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iogann1982 [59]
3 years ago
8

Cancel oh we're on a shelf 1/3 of the cans fell off the show how many did not fall?​

Mathematics
1 answer:
LenaWriter [7]3 years ago
4 0

Answer:

2/3

Step-by-step explanation:

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10 4/5+(8 3/4−6 1/2) no files plz you dont have to explain it either
umka21 [38]

Answer:

11

Step-by-step explanation:

10×4/5+(8×3/4-6×1/2)=8+6-3=14-3=11

6 0
2 years ago
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
Find UV (Pic Stated Below)
natta225 [31]

Answer:

UV=29

Step-by-step explanation:

In right triangles AQB and AVB,

∠AQB = ∠AVB ...(i)  {Right angles}

∠QBA = ∠VBA ...(ii)  {Given that they are equal}

We know that sum of all three angles in a triangle is equal to 180 degree. So wee can write sum equation for each triangle


∠AQB+∠QBA+∠BAQ=180 ...(iii)

∠AVB+∠VBA+∠BAV=180 ...(iv)


using (iii) and (iv)

∠AQB+∠QBA+∠BAQ=∠AVB+∠VBA+∠BAV

∠AVB+∠VBA+∠BAQ=∠AVB+∠VBA+∠BAV  (using (i) and (ii))

∠BAQ=∠BAV...(v)


Now consider triangles AQB and AVB;

∠BAQ=∠BAV  {from (v)}

∠QBA = ∠VBA {from (ii)}

AB=AB  {common side}

So using ASA, triangles AQB and AVB are congruent.

We know that corresponding sides of congruent triangles are equal.

Hence

AQ=AV

5x+9=7x+1

9-1=7x-5x

8=2x

divide both sides by 2

4=x

Now plug value of x=4 into UV=7x+1

UV=7*4+1=28+1=29

<u>Hence UV=29 is final answer.</u>

7 0
3 years ago
10 is multiplied by the sum of 3 and 2. The product is divided by 2, and then 5 is added in numerical expression
bija089 [108]
30 is your answer to this problem

7 0
2 years ago
Read 2 more answers
What is the area of a circle that has a diameter of 18"?
lubasha [3.4K]
Since the formula for the area of a circle is Pi x r squared, the area would be 81 x Pi (you teacher may want you to use Pi as 3.14 or put it in your calculator).
7 0
3 years ago
Read 2 more answers
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