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Goryan [66]
3 years ago
14

When a number is doubled and then three is added, the result is 51. Write an equation, then solve to find the number. Show your

work
Mathematics
1 answer:
Vinil7 [7]3 years ago
3 0

Answer :

Equation :

2y + 3 = 51

Solving :

• 2y + 3 = 51

• 2y = 51 - 3

• 51 - 3 = 48

• 2y = 48

• y = 48 ÷ 2

• 48 ÷ 2 = 24

• so , y = 24

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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
A long distance runner starts at the beginning of a trail and runs at a rate of 4 miles per hour. One hour later, a cyclist star
Scrat [10]
T = 1/4 hour will be the answer
7 0
3 years ago
What are the solutions to the following system of equations?
Cloud [144]
<span>(2, 4) and (−8, 84)
see the graph</span>

4 0
3 years ago
-1/2 + 2/3 DO THIS THANKS GUYSSSSS
BaLLatris [955]

Answer: 0/5

Step-by-step explanation:

8 0
11 months ago
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Can someone please please help me on this algebra problem I’m stuck on answering the question?
Rudik [331]

Explanation:

Midline: it is a horizontal line where half of the function above it and half of the function is below it.

Amplitude: It means how far the function varies from the midline to the above and below.

Period: the distance between two consecutive maximum points (minimum points).

3) From the graph,

\begin{gathered} \text{Midline}=\frac{Maximum\text{ point+ minimum point}}{2} \\ =\frac{3+(-1)}{2} \\ =\frac{3-1}{2} \\ =\frac{2}{2} \\ =1 \end{gathered}

Answer: Midline=1

1) Amplitude

\text{Amplitude =}2

2) Periods

\begin{gathered} \text{Period}=6 \\ \Rightarrow b=\frac{2\pi}{6} \end{gathered}

4) The trigonometric equation is,

\begin{gathered} y=a\cos (b(x-c))+d \\ a=\text{ amplitude} \\ b=\text{period} \\ c=\text{ phase shift} \\ d=\text{ vertical shift} \\ y=2\cos (\frac{2\pi}{6}x)+1 \end{gathered}

Answer:

y=2\cos (\frac{2\pi}{6}x)+1

4 0
1 year ago
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