Write the number that is 10,000 greater than 563262

A rectangular prism is 7 centimeters long, 5 centimeters wide, and 4 centimeters tall. Which values are the areas of cross secti

VladimirAG [237]

Cross section is the area you make when you make a cut across a shape. In this case, the cut must be parallel to the base. Hence, the cut shape would be same to that of the base. The base's dimensions are 7 cm long x 5 cm wide. Thus, the cross section area is 35 square centimeters.

3 0

3 years ago

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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 $ax_1+bx_2$ 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

If y varies directly as x and y=3 when x=2 find y when x=8

satela [25.4K]

Answer:

12

Step-by-step explanation:

x=2 was multiplied by 4 to get to x=8 so you multiply y=3 by 4 to make it y=12

6 0

3 years ago

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A grocer's shop sells carrots, onions and cucumbers.

kolezko [41]

Answer: 3:4

Step-by-step explanation:

7 0

3 years ago

Using the diagram, calculate the values of the unknown variables: a, b, and c.

Lady_Fox [76]

Answer:

a = 82

b = 130

c = 118

Step-by-step explanation:

A quadrilateral is inscribed in a circle. So, it is a cyclic quadrilateral.

Opposite angles of a cyclic quadrilateral are supplementary.

Therefore,

a° + 98° = 180°

a° = 180° - 98°

a° = 82°

a = 82

By inscribed angle theorem:

a° = 1/2(b° + 34°)

82° * 2 = b° + 34°

164° - 34° = b°

b° = 130°

b = 130

Again by inscribed angle theorem:

76° = 1/2(c° + 34°)

76° *2 = c° + 34°

152° - 34° = c°

c° = 118°

c = 118

3 0

3 years ago