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natali 33 [55]
3 years ago
7

Find tue volume of following

Mathematics
1 answer:
kompoz [17]3 years ago
4 0
Whereas the basic formula for the area shape is length x width the basic formula for volume is length x width c height
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Find the direction cosines and direction angles of the vector. (Give the direction angles correct to the nearest degree.) 5, 1,
Dahasolnce [82]

Answer:

The direction cosines are:

\frac{5}{\sqrt{42} }, \frac{1}{\sqrt{42} }  and  \frac{4}{\sqrt{42} }  with respect to the x, y and z axes respectively.

The direction angles are:

40°,  81° and  52° with respect to the x, y and z axes respectively.

Step-by-step explanation:

For a given vector a = ai + aj + ak, its direction cosines are the cosines of the angles which it makes with the x, y and z axes.

If a makes angles α, β, and γ (which are the direction angles) with the x, y and z axes respectively, then its direction cosines are: cos α, cos β and cos γ in the x, y and z axes respectively.

Where;

cos α = \frac{a . i}{|a| . |i|}               ---------------------(i)

cos β = \frac{a.j}{|a||j|}               ---------------------(ii)

cos γ = \frac{a.k}{|a|.|k|}             ----------------------(iii)

<em>And from these we can get the direction angles as follows;</em>

α =  cos⁻¹ ( \frac{a . i}{|a| . |i|} )

β = cos⁻¹ ( \frac{a.j}{|a||j|} )

γ = cos⁻¹ ( \frac{a.k}{|a|.|k|} )

Now to the question:

Let the given vector be

a = 5i + j + 4k

a . i =  (5i + j + 4k) . (i)

a . i = 5         [a.i <em>is just the x component of the vector</em>]

a . j = 1            [<em>the y component of the vector</em>]

a . k = 4          [<em>the z component of the vector</em>]

<em>Also</em>

|a|. |i| = |a|. |j| = |a|. |k| = |a|           [since |i| = |j| = |k| = 1]

|a| = \sqrt{5^2 + 1^2 + 4^2}

|a| = \sqrt{25 + 1 + 16}

|a| = \sqrt{42}

Now substitute these values into equations (i) - (iii) to get the direction cosines. i.e

cos α = \frac{5}{\sqrt{42} }

cos β =  \frac{1}{\sqrt{42} }              

cos γ =  \frac{4}{\sqrt{42} }

From the value, now find the direction angles as follows;

α =  cos⁻¹ ( \frac{a . i}{|a| . |i|} )

α =  cos⁻¹ ( \frac{5}{\sqrt{42} } )

α =  cos⁻¹ (\frac{5}{6.481} )

α =  cos⁻¹ (0.7715)

α = 39.51

α = 40°

β = cos⁻¹ ( \frac{a.j}{|a||j|} )

β = cos⁻¹ ( \frac{1}{\sqrt{42} } )

β = cos⁻¹ ( \frac{1}{6.481 } )

β = cos⁻¹ ( 0.1543 )

β = 81.12

β = 81°

γ = cos⁻¹ ( \frac{a.k}{|a|.|k|} )

γ = cos⁻¹ (\frac{4}{\sqrt{42} })

γ = cos⁻¹ (\frac{4}{6.481})

γ = cos⁻¹ (0.6172)

γ = 51.89

γ = 52°

<u>Conclusion:</u>

The direction cosines are:

\frac{5}{\sqrt{42} }, \frac{1}{\sqrt{42} }  and  \frac{4}{\sqrt{42} }  with respect to the x, y and z axes respectively.

The direction angles are:

40°,  81° and  52° with respect to the x, y and z axes respectively.

3 0
3 years ago
Sin θ = 12/37. Find tan θ.
ivann1987 [24]

Answer:

B

Step-by-step explanation:

Use SOHCAHTOA.

sinθ = \frac{opposite}{hypotenuse}

Opposite side length= 12

Hypotenuse length = 37

Find the adjacent side length (b) using the Pythagorean theorem.

12^2 + b^2 = 37^2

b = 35

tanθ = \frac{opposite}{adjacent} = 12/35

Choice B is the correct answer.

6 0
3 years ago
Read 2 more answers
Carson has two cans of red paint. The first can contains 3 pints of red paint. The second can contains 4 16 pints of red paint.
allsm [11]

Answer: 1⁵/₈ pints

Step-by-step explanation:

First can contained 3¹/₄ pints and second can contained 6¹/₂.

To make them equal, first find the different;

= 6¹/₂ - 3¹/₄

= 6.5 - 3.25

= 3.25

Divide this difference by 2;

= 3.25/2

= 1.625

= 1⁶²⁵/₁₀₀₀

= 1⁵/₈ pints

If Carson pours 1⁵/₈ pints from the second Can, it would increase the first can to 4⁷/₈ pints whilst reducing the second can to 4⁷/₈ pints as well.

8 0
3 years ago
A computer and printer cost a total of $900. The cost of the computer is three times the cost of the printer. Find the cost of e
KiRa [710]

Answer:

C is computer

P is printer

3c + p = 900

750 + 150 = 900

So...

Computer = $750

Printer = $150

Brainiest plz :)

6 0
3 years ago
Use the Law of Sines to find the measure of angle J to the nearest degree.
rjkz [21]
\bf \textit{Law of sines}&#10;\\ \quad \\&#10;\cfrac{sin(\measuredangle A)}{a}=\cfrac{sin(\measuredangle B)}{b}=\cfrac{sin(\measuredangle C)}{c}\\\\&#10;-----------------------------\\\\&#10;\cfrac{sin(J)}{9.1}=\cfrac{sin(97^o)}{11}\implies sin(J)=\cfrac{9.1\cdot sin(97^o)}{11}&#10;\\\\\\&#10;\textit{now taking }sin^{-1}\textit{ to both sides}&#10;\\\\\\&#10;sin^{-1}\left[ sin(J) \right]=sin^{-1}\left( \cfrac{9.1\cdot sin(97^o)}{11} \right)&#10;\\\\\\&#10;\measuredangle J=sin^{-1}\left( \cfrac{9.1\cdot sin(97^o)}{11} \right)
8 0
4 years ago
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