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maw [93]
3 years ago
9

What is the standard form polynomial representing the volume of this shipping container?

Mathematics
1 answer:
liq [111]3 years ago
6 0

Answer:

Step-by-step explanation:

Volume of the shipping container = Lengt * Breadth * Height

Given

Length = 4x² + 3x

Breadth = x² – 8

Height = 6x +15

Volume of the container = ( 4x² + 3x)( x² - 8)(6x+15)

( 4x² + 3x)( x² - 8)  = 4x⁴-32x²+3x³-24x

( 4x² + 3x)( x² - 8) = 4x⁴+3x³-32x²-24x

(4x⁴+3x³-32x²-24x)(6x+15) = 24x⁵+60x⁴+18x⁴+45x³-192x³-480x²-144x²-360x

Collect like terms

( 4x² + 3x)( x² - 8)(6x+15) = 24x⁵+78x⁴-147x³-624x²-360x

Hence the standard form polynomial representing the volume of this shipping container is expressed as V = 24x⁵+78x⁴-147x³-624x²-360x

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4 0
3 years ago
Terry needs to carry a pole that
Natasha2012 [34]

Answer:

Yes, The pole will fit through the door because the diagonal width of the door is 10.8 feet, which is longer than the length of the pole.

Step-by-step explanation:

Using the Pythagorean Theorem, (a^2+b^2=c^2 ) we can measure the hypotenuse of a right triangle. Since the doorway is a rectangle, and a rectangle cut diagonally is a right triangle, we can use Pythagorean Theorem to measure the diagonal width of the doorway.

Plug in the values of the length and width of the door for a and b. The c value will represent the diagonal width of the doorway:

6^2+9^2=c^2

36+81=c^2

117=c^2

Since 117 is equal to the value of c multiplied by c, we must find the square root of 117 to find the value of c.

\sqrt{117} =10.8

10.8=c

Yes, The pole will fit through the door because the diagonal width of the door is 10.8 feet, which is longer than the length of the pole, measuring 10 feet.

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What does P(X4=2,X3=3,X2=3,X1=2)
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3 years ago
Given tan theta =9, use trigonometric identities to find the exact value of each of the following:_______
Ludmilka [50]

Answer:

(a)\ \sec^2(\theta) = 82

(b)\ \cot(\theta) = \frac{1}{9}

(c)\ \cot(\frac{\pi}{2} - \theta) = 9

(d)\ \csc^2(\theta) = \frac{82}{81}

Step-by-step explanation:

Given

\tan(\theta) = 9

Required

Solve (a) to (d)

Using tan formula, we have:

\tan(\theta) = \frac{Opposite}{Adjacent}

This gives:

\frac{Opposite}{Adjacent} = 9

Rewrite as:

\frac{Opposite}{Adjacent} = \frac{9}{1}

Using a unit ratio;

Opposite = 9; Adjacent = 1

Using Pythagoras theorem, we have:

Hypotenuse^2 = Opposite^2 + Adjacent^2

Hypotenuse^2 = 9^2 + 1^2

Hypotenuse^2 = 81 + 1

Hypotenuse^2 = 82

Take square roots of both sides

Hypotenuse =\sqrt{82}

So, we have:

Opposite = 9; Adjacent = 1

Hypotenuse =\sqrt{82}

Solving (a):

\sec^2(\theta)

This is calculated as:

\sec^2(\theta) = (\sec(\theta))^2

\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2

Where:

\cos(\theta) = \frac{Adjacent}{Hypotenuse}

\cos(\theta) = \frac{1}{\sqrt{82}}

So:

\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2

\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2

\sec^2(\theta) = (\sqrt{82})^2

\sec^2(\theta) = 82

Solving (b):

\cot(\theta)

This is calculated as:

\cot(\theta) = \frac{1}{\tan(\theta)}

Where:

\tan(\theta) = 9 ---- given

So:

\cot(\theta) = \frac{1}{\tan(\theta)}

\cot(\theta) = \frac{1}{9}

Solving (c):

\cot(\frac{\pi}{2} - \theta)

In trigonometry:

\cot(\frac{\pi}{2} - \theta) = \tan(\theta)

Hence:

\cot(\frac{\pi}{2} - \theta) = 9

Solving (d):

\csc^2(\theta)

This is calculated as:

\csc^2(\theta) = (\csc(\theta))^2

\csc^2(\theta) = (\frac{1}{\sin(\theta)})^2

Where:

\sin(\theta) = \frac{Opposite}{Hypotenuse}

\sin(\theta) = \frac{9}{\sqrt{82}}

So:

\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2

\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2

\csc^2(\theta) = \frac{82}{81}

4 0
3 years ago
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