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andrezito [222]
3 years ago
14

A container full of water weighs 15.5 lbs. When ¼ of the water is poured out, the container now weighs 12 ½ lbs. How much does t

he container weigh when empty?
Mathematics
1 answer:
posledela3 years ago
5 0

Answer:

3.5 lbs.

Step-by-step explanation:

1/4 of the water is dumped, meaning that is 3 lbs. 1/4 is 3 lbs. 3x4 is 12. 15.5 - 12 = 3

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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
Find the coordinates of the midpoint AB. 1) ( -2,4), (4,0) THANK YOU!! And it's worth a lot of points if I get it right.
AnnyKZ [126]
-2+4 is 2/2 is (1,y)
4+0 divided by 2 is 2 
coordinate is (1,2)
3 0
3 years ago
What is the answer I need help ASAP
ira [324]

Answer:

C) 4

Step-by-step explanation:

5 0
3 years ago
If a cross-country runner runs a 55 kilometer (km) race, how many miles is that? Round your answer to one decimal place.
prohojiy [21]

Answer: 34.2

For everyone 1 kilometer=.621371 so if you times that by 55 you get 34.1754 which rounds to 34.2

5 0
3 years ago
F is the midpoint of GH. Point G is located at 32 and F is at 105. What is the location of point H?
sukhopar [10]
The location of point H is 68.
5 0
3 years ago
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