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3 years ago

The President's cabinet is best described as a:

Mathematics

2 answers:

svet-max [94.6K]3 years ago

5 0

Answer:

number 3

Step-by-step explanation:

Kisachek [45]3 years ago

5 0

Answer:

It's actually number 2

Step-by-step explanation:

I took the test

You might be interested in

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

Which equation involves a prime quadratic and cannot be solved by factoring? A. x2 + 5x − 4 = 0 B. x2 − x − 6 = 0 C. x2 + 3x − 4

Damm [24]

Answer:

Option A

Step-by-step explanation:

We can try to factorize all the options one by one.

For A:

$x^{2} +5x-4\\=> x^{2}+4x+x-4\\=>x(x+4)+1(x-4)$

We can see that the quadratic expression cannot be solved by factorization as the factors at the end of factorization are not equal in both brackets. So Option A is the correct answer for the given question.

Moreover we can also note that all the other quadratic expressions can be factorized ..

8 0

3 years ago

Mario’s company makes unusually shaped imitation gemstones. One gemstone has 14 faces and 10 vertices. How many edges does the g

Ivan

The gemstone has 22 edges.

We use Euler's formula on this:

F + V - E = 2

Substituting our information, we have:

14 + 10 - E = 2

Combining like terms,
24 - E = 2

Subtract 24 from both sides:
24 - E - 24 = 2 - 24
-E = -22

Divide both sides by -1:
-E/-1 = -22/-1
E = 22

3 0

3 years ago

At 19:30 Jack lights a campfire. At 22:15 he puts the fire out. Yeah

Evgen [1.6K]

Yeah because he Sheldon do I be j I got work y’all got work at

3 0

3 years ago

Which is the completely factored form of 12x^3-60x^2+4x-20

Shkiper50 [21]

Answer:

(12x²+4)(x-5)

Step-by-step explanation:

To factor the expression factor by grouping.

(12x³ - 60x²) + (4x - 20)

12x²(x - 5) 4(x-5)

(12x²+4)(x-5)

6 0

3 years ago