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777dan777 [17]
2 years ago
9

Which is true about the completely simplified differences of the polynomials a^3b+9a^2b^2-4ab^5 and a^3b-3a^2b^2+ab^5 ?

Mathematics
2 answers:
tino4ka555 [31]2 years ago
4 0

Answer:

Option D.

Step-by-step explanation:

The given polynomials are a^{3}b+9a^{2}b^{2}-4ab^{5} and a^{3}b-3a^{2}b^{2}+ab^{5}

Now we will subtract the 1st polynomial from second.

a^{3}b+9a^{2}b^{2}-4ab^{5} - (a^{3}b-3a^{2}b^{2}+ab^{5})

= a^{3}b-a^{3}b+9a^{2}b^{2}+3a^{2}b^{2}-4ab^{5}-ab^{5}

= 12a^{2}b^{2}-5ab^{5}

Now the degree of both the terms of the polynomial is

Degree of 1st term = 2 + 2 = 4

Degree of 2nd term = 1 + 5 = 6

Therefore, the highest degree of the polynomial is 6.

Option D. will be the answer.

dolphi86 [110]2 years ago
3 0

Hello from MrBillDoesMath!

Answer:

Binomial with a degree of 6  (the second Choice)

Discussion:

(a^3b+9a^2b^2-4ab^5)  - (a^3b-3a^2b^2+ab^5) =

(-4ab^5- ab^5) +  ( a^3b-a^3b) + ( 9a^2b^2 + 3a^2b^2) =

(-4ab^5- ab^5) +  0                   + ( 9a^2b^2 + 3a^2b^2) =

12 a^2 b^2 - 5 a b^5

This is a binomial with degree 6 (degree of last term = 1 + 5 = 6).

Thank you,

MrB

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nevsk [136]

For given f(x, y) the extremum: (12, 24) which is the minimum.

For given question,

We have been given a function f(x) = 4x² + 2y² under the constraint 3x+3y= 108

We use the constraint to build the constraint function,

g(x, y) = 3x + 3y

We then take all the partial derivatives which will be needed for the Lagrange multiplier equations:

f_x=8x

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Setting up the Lagrange multiplier equations:

f_x=\lambda g_x

⇒ 8x = 3λ                                        .....................(1)

f_y=\lambda g_y

⇒ 4y = 3λ                                         ......................(2)

constraint: 3x + 3y = 108                .......................(3)

Taking (1) / (2), (assuming λ ≠ 0)

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Substitute this value of y in equation (3),

⇒ 3x + 3y = 108

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Now we find the value of f(12, 24)

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From (1) and (2),

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Learn more about the extremum here:

brainly.com/question/17227640

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