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3 years ago

Arrange the terms of the polynomial in descending powers of b 16-8b^7+5b-4b^5

Mathematics

1 answer:

maria [59]3 years ago

6 0

Answer:

-8b^7 - 4b^5 + 5b + 16

Step-by-step explanation:

The powers are indicated by the value of the exponents, so move top down.

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PLEASE HURRY ITS TIMED 20 POINTS!!!What is the y-coordinate of the point shown in the graph?

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<h3><em>X CO-ORDINATE = 3</em></h3><h3><em>Y CO-ORDINATE = -7</em></h3><h3><em>HOPE IT HELPS.....BYE....</em></h3>

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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 $R^3$

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3 years ago

How would you translate this situation below into an algebraic expression?

Natasha2012 [34]

Dime value: $0.10
Nickel value: $0.05

For this word problem, you are asked to create an algebraic expression (an equation) for the unknown variables.

Now, we know that a dime is worth $0.10 or 10 cents and a nickel is worth $0.05 or 5 cents. So, all you would have to do is create an equation that takes those values into account.
5n + 10d

Your algebraic expression would be B. 5n + 10d

I hope this helps!

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3 years ago

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3 years ago

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Vladimir [108]

Answer:

a) (x+2)(x+3)

b) (x+2)(x+3)=0, x=-2 and x=-3

Step-by-step explanation:

3 0

3 years ago

Arrange the terms of the polynomial in descending powers of b 16-8b^7+5b-4b^5​

Arrange the terms of the polynomial in descending powers of b 16-8b^7+5b-4b^5