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FromTheMoon [43]
3 years ago
9

Which polynomial correctly combines the like terms and puts the given polynomial in standard form? -5x^3y^3+8x^4y^2-xy^5-2x^2y^4

+8x^6+3x^2y^4-6xy^5
Mathematics
1 answer:
victus00 [196]3 years ago
8 0

Given:

The polynomial is:

-5x^3y^3+8x^4y^2-xy^5-2x^2y^4+8x^6+3x^2y^4-6xy^5

To find:

The simplified form of the given polynomial in standard form.

Solution:

We have,

-5x^3y^3+8x^4y^2-xy^5-2x^2y^4+8x^6+3x^2y^4-6xy^5

Combining the like terms, we get

=-5x^3y^3+8x^4y^2+(-2x^2y^4+3x^2y^4)+8x^6+(-xy^5-6xy^5)

=-5x^3y^3+8x^4y^2+x^2y^4+8x^6-7xy^5

Now, rewrite this polynomial in standard form.

=8x^6+8x^4y^2-5x^3y^3+x^2y^4-7xy^5

Therefore, the required polynomial in standard form is 8x^6+8x^4y^2-5x^3y^3+x^2y^4-7xy^5.

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If tan a = 9/40<br>use trigonometric identities to find the values of sin a and cos a.​
Sindrei [870]

Answer:

see explanation

Step-by-step explanation:

Given

tan a = \frac{9}{40} = \frac{opposite}{adjacent}

We require the hypotenuse h

Using Pythagoras' identity

h² = 9² + 40² = 81 + 1600 = 1681 ( take square root of both sides )

h = \sqrt{1681} = 41 , thus

sin a = \frac{opposite}{hypotenuse} = \frac{9}{41}

cos a = \frac{adjacent}{hypotenuse} = \frac{40}{41}

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An emperor penguin is 45 inches tall. It is 24 inches taller than a rockhopper penguin. Write and solve an equation to find the
Vlad [161]
H= height of rockhopper penguin.
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A large storage tank, open to the atmosphere at the top and filled with water, develops a small hole in its side at a point 10.4
vlabodo [156]

Answer:

Therefore the diameter of the hole is 1.94 \times 10^{-3} m.

Step-by-step explanation:

Bernoulli's equation,

P_1+\frac12 \rho v^2_1+\rho g h_1= P_2+\frac12 \rho v^2_2+\rho g h_2

P₁ = P₂= atmospheric presser

\rho= density

\frac12 \rho v^2_1+\rho g h_1= \frac12 \rho v^2_2+\rho g h_2             [since P₁ = P₂]

\Rightarrow\rho (\frac12 v^2_1+ g h_1)= \rho(\frac12 v^2_2+ g h_2)

\Rightarrow\frac12 v^2_1+ g h_1= \frac12 v^2_2+ g h_2

\Rightarrow\frac12 v^2_2-\frac12 v^2_1=g h_1- g h_2

\Rightarrow v^2_2- v^2_1=2g h                                [h_1-h_2=h]

Here   v_1\approx 0

\Rightarrow v^2_2=2g h

\therefore v_2=\sqrt {2gh

Here g= 9.8 m/s² , h = 10.4 m

The velocity of water that leaves from the hole v_2 = \sqrt {2\times 9.8\times 10.4} m/s

                                                                                  =14.28 m/s.

Given, the rate of flow from the leak is 2.53\times 10^{-3} m^3/min

                                                               =\frac{2.53\times 10^{-3}}{60}  m^3/s

Let the diameter of the hole be d.

Then the cross section area of the hole is =\pi (\frac d2)^2

We know that,

The rate of flow = Cross section area × speed

\Rightarrow \frac{2.53\times 10^{-3}}{60} =\pi (\frac d2)^2\times 14.28

\Rightarrow (\frac d2)^2=\frac{2.53\times 10^{-3}}{60\times 14.28\times \pi}

\Rightarrow d= 1.94 \times 10^{-3}

Therefore the diameter of the hole is 1.94 \times 10^{-3} m.

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strojnjashka [21]

Answer:

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